Local non-intrinsic property of surface

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Weingarten equations

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Weingarten equations

Weingarten equations is about derivative of the unit normal vector \( \vec{N} \).
These equations were established in 1861 by German mathematician Julius Weingarten.
The Weingarten equation is
\( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
\( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)
The matrix equivalent of the Weingarten equation is
\( {H^2} \begin{pmatrix} \vec{N}_1 \\ \vec{N}_2 \end{pmatrix}=\begin{pmatrix} FM-GL & FL-EM \\ FN-GM & FM-EN \end{pmatrix} \begin{pmatrix} \vec{r}_1 \\ \vec{r}_2 \end{pmatrix} \)

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In a surface, show that

1. \( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
2. \( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)

Also, verify that
\( H\vec{N}_1\times \vec{N}_2=( LN-{M^2} )\vec{N}\)
Proof
Let \( S:\vec{r}=\vec{r}( u,v )\) be a surface and \( \vec{N}\) be unit normal.
Then,
\( d\vec{N}\) lies in tangent plane of S. Thus, we write
\( \vec{N}_1=a\vec{r}_1+b\vec{r}_2\) (A)
\( \vec{N}_2=c\vec{r}_1+d\vec{r}_2\) (B)
Taking dot product on both sides of (A) by \( \vec{r}_1\) we get
\( \vec{N}_1.\vec{r}_1=a\vec{r}_1^2+b\vec{r}_1.\vec{r}_2\)
or \( -L = aE+bF \)
or \( aE+b F + L =0 \) (1)
Again,
Taking dot product on both sides of (A) by \( \vec{r}_2\) we get
\( \vec{N}_1.\vec{r}_2=a\vec{r}_1.\vec{r}_2+b\vec{r}_2^2\)
or \( -M = a F +b G \)
or \( aF+bG+M=0 \) (2)
Now,
The cross-multiplication method is applied.
Solving, equations (1) and (2), for \( a \) and \( b \) ,we get
\( \frac{a}{ FM-GL } = \frac{b}{ FL-EM }= \frac{1}{EG-F^2}\)
or \( \frac{a}{ FM-GL } = \frac{b}{ FL-EM }= \frac{1}{H^2}\)
Equating first and third identity, similarly equating second and third identity, we get
or \( a=\frac{ FM-GL }{ H^2} , b= \frac{ FL-EM }{ H^2}\)
Substituting, value of \( a \) and \( b\) in (A) we get
\( \vec{N}_1=a\vec{r}_1+b\vec{r}_2\)
or \( \vec{N}_1=\frac{ FM-GL }{ H^2} \vec{r}_1+\frac{ FL-EM }{ H^2} \vec{r}_2\)
or \( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\)
The first Weingarten equation established.

Similarly,
Taking dot product on both sides of (B) by \( \vec{r}_1\) we get
\( \vec{N}_2.\vec{r}_1=c\vec{r}_1^2+d\vec{r}_1.\vec{r}_2\)
or \( -M = cE+dF \)
or \( cE+d F + M =0 \) (3)
Again,
Taking dot product on both sides of (B) by \( \vec{r}_2\) we get
\( \vec{N}_2.\vec{r}_2=c\vec{r}_1.\vec{r}_2+d\vec{r}_2^2\)
or \( -N = c F +d G \)
or \( cF+dG+N=0 \) (4)
Now,
Solving, equations (3) and (4), for \( c \) and \( d \) ,we get
The cross-multiplication method is applied.
\( \frac{c}{ FN-GM } = \frac{d}{ FM-EN }= \frac{1}{EG-F^2}\)
or \( \frac{c}{ FN-GM } = \frac{d}{ FM-EN }= \frac{1}{H^2}\)
Equating first and third identity, similarly equating second and third identity, we get
\( c=\frac{ FN-GM }{ H^2} , d= \frac{ FM-EN }{ H^2}\)
Substituting, value of \( c \) and \(d\) in (B) we get
\( \vec{N}_2=c\vec{r}_1+d\vec{r}_2\)
or \( \vec{N}_2=\frac{ FN-GM }{ H^2} \vec{r}_1+\frac{ FM-EN }{ H^2} \vec{r}_2\)
or \( H^2 \vec{N}_2=( FN-GM ) \vec{r}_1+( FM-EN ) \vec{r}_2\)
The second Weingarten equation established.

Finally,
The Weingarten equation is
\( {H^2}\vec{N}_1=( FM-GL )\vec{r}_1+( FL-EM )\vec{r}_2\) and
\( {H^2}\vec{N}_2=( FN-GM )\vec{r}_1+( FM-EN )\vec{r}_2\)
Taking cross product of Weingarten equations, we get
\( {H^2}\vec{N}_1 \times {H^2}\vec{N}_2 = \left\{ ( FM-GL )\vec{r}_1 + ( FL-EM )\vec{r}_2 \right \} \times \left \{ ( FN-GM )\vec{r}_1 + ( FM-EN )\vec{r}_2 \right \} \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL )\vec{r}_1 \times ( FM-EN ) \vec{r}_2 + (FL-EM )\vec{r}_2 \times ( FN-GM )\vec{r}_1 \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) (\vec{r}_1 \times \vec{r}_2 )+ (FL-EM ) ( FN-GM ) ( \vec{r}_2 \times \vec{r}_1) \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) ( \vec{r}_1 \times \vec{r}_2 ) - (FL-EM ) ( FN-GM ) (\vec{r}_1 \times \vec{r}_2) \)
or\( H^4\vec{N}_1 \times \vec{N}_2 = ( FM-GL ) ( FM-EN ) H \vec{N} - (FL-EM ) ( FN-GM ) H \vec{N} \)
or\( H^4 \vec{N}_1 \times \vec{N}_2 = \{ ( FM-GL ) ( FM-EN ) - (FL-EM ) ( FN-GM ) \} H\vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = ( EG-F^2 ) ( LN-M^2 ) H \vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = H^2 ( LN-M^2 ) H \vec{N} \)
or\( {H^4}\vec{N}_1 \times \vec{N}_2 = H^3 ( LN-M^2 ) \vec{N} \)
or\( H\vec{N}_1 \times \vec{N}_2 = ( LN-M^2 ) \vec{N} \)
This completes.

Prove that, a oint \( p \) on a Surface is Umbilical if and only if \( \mu^2 = K \)
We know that
A point \( p \) on a surface is called umbilical if and only if the principal curvatures at \( p \) are equal, i.e., \( k_1 = k_2 \). So, If \( p \) is umbilical, then \(\mu = \frac{k_1 + k_2}{2} = \frac{k + k}{2} = k\)
Next
\(K = k_1 k_2 = k \times k = k^2\)
This shows that if \( p \) is an umbilical point, then \( \mu^2 = K \)

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Double family of curves

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Double family of curves

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface then the quadratic differential equation
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)^2+2Q\left( \frac{du}{dv}\right)+R=0 \)
where P,Q,R are continuous function of u,v and do not vanish together, is called double family of curves on the surface.

Condition for orthogonality of double family

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface, and
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)+2Q\left( \frac{du}{dv}\right)+R=0 \)
be double family of curves.
If \(\left( \frac{\lambda}{\mu}, \frac{\lambda '}{\mu '}\right) \) be the roots of the curves, then.
\( \frac{\lambda}{\mu}+ \frac{\lambda '}{\mu '}= \frac{-2Q}{P}\) and \( \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}= \frac{R}{P}\)
Now, the curves are orthogonal if
\( E \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}+ F\left( \frac{\lambda}{\mu}+\frac{\lambda '}{\mu '}\right)+G=0\)
or\( E \left ( \frac{R}{P} \right ) + F \left (\frac{-2Q}{P} \right )+G=0\)
or\( ER-2QF+GP=0\)

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Theorem 1

The necessary and sufficient condition for parametric curves to be orthogonal is F=0
Proof
Let \(S: \vec{r}=\vec{r}(u,v)\) be a surface with parametric curves u =constant and v =constant on it.
Then, differential equation of parametric curves is
dudv=0 (i)
Also, differential equation of double family of curves is
\(P{du}^2+2Qdudv+R{dv}^2=0\) (ii)
Comparing (i) and (ii) we get
P=0, Q≠0 and R=0
Now, necessary and sufficient condition for parametric curves to be orthogonal is
ER-2QF+GP=0
or E.0-2QF+G.0=0
or F=0

Theorem 2

Show that parametric curves form an orthogonal system on a sphere x=asinu cosv,y=asinu sinv,z=acosu.
Solution
The sphere is
x=asinucosv,y=asinusinv,z=acosu
or \( \vec{r}=(a \sin u \cos v, a \sin u \sin v,a \cos u \) (i)
By successive differentiation w. r. to. u and v, we get
\( \vec{r}_1=(a \cos u \cos v,a \cos u \sin v,-a \sin u \)
\( \vec{r}_2=(-a \sin u \sin v,a \sin u \cos v,0 \)
Now, the fundamental coefficient are
\( F=\vec{r}_1.\vec{r}_2=0\)
Hence, the parametric curves on a sphere form an orthogonal system.

Theorem 3

Prove that, if θ is angle between two directions of \(Pdu^2+2Qdudv+Rdv^2=0\) then \( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
Solution
The double family of curves is
\(Pdu^2+2Qdudv+Rdv^2=0\)
or \(P \left ( \frac{du}{dv} \right )^2+2Q \frac{du}{dv}+R=0\)
Let \( \left ( \frac{l}{m} ,\frac{l'}{m'} \right ) \) be the roots, then
\( \frac{l}{m}+\frac{l'}{m'}=\frac{-2Q}{P}\) and \( \frac{l}{m}.\frac{l'}{m'}=\frac{R}{P}\)
Hence
\( \cos \theta =Ell'+F(lm'+l'm)+Gmm'\)
or \( \cos \theta =E \left( \frac{l}{m}.\frac{l'}{m'}\right )+F \left( \frac{l}{m}+\frac{l'}{m'}\right )+G\)
or \( \cos \theta =E \left( \frac{R}{P} \right ) +F \left( \frac{-2Q}{P} \right )+G\)
or \( \cos \theta =\frac{ER-2FQ+GP}{P} \) (A)
Again
\( \sin \theta =H(lm'-l'm)\)
or \( \sin \theta =H\left( \frac{l}{m}-\frac{l'}{m'}\right )\)
or \( \sin \theta =H \sqrt{\left( \frac{l}{m}+\frac{l'}{m'}\right )^2-4 \left( \frac{l}{m}.\frac{l'}{m'}\right ) } \)
or \( \sin \theta =H \sqrt{\left( \frac{-2Q}{P} \right )^2-4 \left( \frac{R}{P} \right ) } \)
or \( \sin \theta =\frac{2H \sqrt{Q^2-PR}}{P} \) (B)
Thus, using (A) and (B), we get
\( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
This completes the solution

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Orthogonal trajectories

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Orthogonal trajectories

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface with family of curves
\( \phi(u,v)=c\) (i)
\( \psi(u,v)=c_1\) (ii)
Now, family (i) and (ii) are called orthogonal trajectories, if the curves of the both families are orthogonal at each of their intersection.

Examples

As we know, family of curves \( xy=c;c\neq 0 \) and \( x^2-y^2=c;c\neq 0\) of hyperbolas are orthogonal to each other, thus these families are orthogonal trajectories to each other.
Similarly, \(uv=c;c\neq 0 \) and \( u^2-v^2=c;c\neq 0\) are orthogonal to each other on the surface, so these families are orthogonal trajectories to each other.

Next, the family of circles \(x^2+y^2=c\) and that of lines \(y=mx\) are orthogonal trajectories to each other.
Similarly, \(u^2+v^2=c \) and \( v=mu\) are orthogonal to each other on the surface, so these families are orthogonal trajectories to each other.

Differential equation of orthogonal trajectories

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface. If
\( \phi(u,v)=c\) (i) be a family of curves with directions (du,dv)
\( \psi(u,v)=c_1\) (ii)be another family of curves with directions (-Q,P)
Then, the families (i) and (i) are orthogonal trajectories if (du,dv) and (-Q,P) are orthogonal.
Then, differential equation of orthogonal trajectories is
\(Ell'+F(lm'+l'm))+Gmm'=0\)
or \(Edu(-Q)+F(Pdu-Qdv)+GPdv=0\)
or \((FP-EQ)du+(GP-FQ)dv=0 \)

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Family of Curves

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Family of Curves

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface and
\( \phi(u,v)=c\) (i)
be a single valued function of \((u,v)\) having continuous derivatives \( \phi_1\) and \( \phi_2\) which do not vanish together. Then an equation (i) where c is real parameter gives a family of curves lying on the surface.
For different values of c, (i) gives different curves on the surface.

Example

1. As we know \( x^2+y^2=c\); where c is a real parameter represent a family of circles
   Similarly, \( u^2+v^2=c\); where c is a real parameter represent a family of circles on the surface\)
   
   
   
2. As we know \( y=mx\); where m is real parameter represent a family of straight
   Similarly, \( v=mu\); where m is a real parameter represent a family of curves on the surface
   
   
   
3. As we know, \(xy=c;c \ne 0\); where c is real parameter represent a family hyperbola
   Similarly, \( uv=c\); where c is a real parameter represent a family of curves on the surface
   
   
   
4. As we know, \(x^2-y^2=c;c\neq 0\); where c is real parameter represent another family hyperbolas
   Similarly, \( u^2-v^2=c\); where c is a real parameter represent a family of curves on the surface
   
   
   

Differential equation of family of curves

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface and \( \phi(u,v)=c\) be a family of curves, then
\( \phi(u,v)=c\)
or\( \phi_1 du+\phi_2 dv=0 \)
or\(\frac{du}{dv}=- \frac{\phi_2}{\phi_1} \)
or\(\frac{du}{dv}=- \frac{Q}{P} \) where \((-\phi_2,\phi_1) \) are proportional to (-Q,P)
is called differential equation of family of curves.

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Direction coefficients and related results

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Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface.
Then tangent line to the surface is described by
\( \vec{T}=\lambda \vec{r}_1+\mu \vec{r}_2 \)
Here
\( (\lambda, \mu )\) is called direction components
If \( \vec{e} \) is unit vector along this tangent line, then
\( \vec{e}=l \vec{r}_1+m\vec{r}_2\)
Here
\( (l, m )\) is called direction coefficients
in which
\( El^2+2Flm+Gm^2=1\)

Relation between \( (\lambda,\mu)\) and \( (l,m) \)

Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface with direction components \( (\lambda, \mu )\) and direction coefficients \( (l, m )\) then
\( \frac{l}{\lambda},\frac{m}{\mu} =k \), (say)
or \( l=\lambda k, m= \mu k\)

Now,
\( El^2+2Flm+Gm^2=1\)
or\( E(\lambda k)^2+2F(\lambda k)(\mu k)+G(\mu k)^2=1\)
or\( k^2 (E \lambda ^2+2F \lambda \mu +G \mu^2) =1\)
or \( k=\frac{1}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}}\)
Thus,
\( (l, m )= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)

Direction coefficients of parametric curves

Let S:\(\vec{r}=\vec{r}(u,v)\) be a surface with parametric curves
u=constant (v-curve) and v=constant (u-curve).
Then, \(\vec{r}_1\) is tangent to v= constant curve, thus
\(\vec{r}_1=1\cdot \vec{r}_1+0\cdot \vec{r}_2\)
Here, component of \(\vec{r}_1\) is (1,0), so direction coefficient of v= constant curve (u-curve) is
\((l,m)= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)
or\((l,m)= \left ( \frac{1}{\sqrt{E}},0 \right )\)
Similarly,
we know that, \(\vec{r}_2\) is tangent to u= constant curve (v-curve), thus
\(\vec{r}_2=0 \cdot \vec{r}_1+1 \cdot \vec{r}_2\)
Here, component of \(\vec{r}_2\) is (0,1), so direction coefficient of u= constant curve (v-curve) is
\((l,m)= \left ( \frac{\lambda}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}},\frac{\mu}{\sqrt{E \lambda ^2+2F \lambda \mu +G \mu^2}} \right )\)
or\((l,m)= \left ( 0,\frac{1}{\sqrt{G}} \right )\)

Angle between two directions

Let S: \(\vec{r}=\vec{r}(u,v)\) be a surface \( (l,m)\) and \( (l' ,m')\) be two direction at P, then their corresponding unit vectors are
\( \vec{e} = l\vec{r}_1+m \vec{r}_2 \)
\( \vec{e}' = l' \vec{r}_1+m' \vec{r}_2 \)
If θ be the angle between these directions then
\( \cos \theta = \vec{e}. \vec{e}' \)
or \( \cos \theta = (l\vec{r}_1+m \vec{r}_2) . ( l' \vec{r}_1+m' \vec{r}_2 ) \)
or \( \cos \theta = Ell'+F(lm'+l'm)+Gmm' \)
Also
\( \sin \theta = | \vec{e} \times \vec{e}' | \)
or \( \sin \theta = |(l\vec{r}_1+m \vec{r}_2) \times ( l' \vec{r}_1+m' \vec{r}_2 )| \)
or \( \sin \theta =H(lm'-l'm) \)
Therefore
\( \tan \theta = \frac{H(lm'-l'm)}{Ell'+F(lm'+l'm)+Gmm'} \)
Note
If two directions \( (l,m)\) and \( (l' ,m')\) are orthogonal, then
\( Ell'+F(lm'+l'm)+Gmm'=0 \)
or \( E \frac{l}{m} \frac{l'}{m'}+F \left (\frac{l}{m} + \frac{l'}{m'} \right )+G'=0 \)
Equivalently, if two directions \( (\lambda,\mu)\) and \( (\lambda' ,\mu')\) are orthogonal, then
\( E \lambda \lambda '+F(\lambda \mu '+\lambda ' \mu)+G \mu \mu'=0 \)
or \( E \frac{\lambda}{\mu} \frac{\lambda'}{\mu'}+F \left (\frac{\lambda}{\mu} + \frac{\lambda'}{\mu'} \right )+G'=0 \)

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Fundamental Cofficients of Surface

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Some Common Example of Surface

1. Plane surface

   
   Plane is a surface traced by a straight line whose parameters are of degree 1. One example of plane surface is given by
   \( \vec{r}=(u, v,u+v) \)
   

2. Cylinder

   
   Cylinder is a surface traced by a straight line being parallel to a fixed vector. It is given by an equation
   \( \vec{r}=(r \cos u, r \sin u,v) \)
   

3. Cone

   
   Cone is a surface traced by a straight line being fixed to a fixed point. It is given by an equation
   \( \vec{r}=(v \cos u, v \sin u,v) \)
   

4. Paraboloid

   \(\vec{r}=( u,v,u^2+v^2 ) \)

5. Hyperboloid

   \( \vec{r}=(x,y,x^2-y^2) \)

6. Minimal surface

   \( \vec{r}=(x,y,\log \cos y-\log \cos x) \)

7. The helicoid

   \( \vec{r}=(u \cos v,u \sin v ,v) \)

8. Pseduo-sphere

   \(\vec{r}=(\sec h u \cos v, \sec h u \sin v, u-\tan h )\)

9. Monge’s form

   \( \vec{r}=( u,v,f( u,v ) )\)

10. Surface of revolution

    \(\vec{r}=(u \cos v,u \sin v, f( u ))\)

11. Conoidal surface

    \( \vec{r}=( u \cos v,u \sin v,f( v ) ) \)

12. Saddle surface

    \( \vec{r}=( u,v,uv) \)

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Compute Fundamental Cofficients of surface

Compute fundamental coefficients for a saddle surface \( \vec{r}=(u,v,uv) \)
Solution
The saddle surface is
\( \vec{r}=(u,v,uv) \)(i)
Differentiation of (i) w. r. to. u and v, we get
\( \vec{r}_1=(1,0,v) \)
\( \vec{r}_2=(0,1,u) \)
\( \vec{r}_{11}=(0,0,0) \)
\( \vec{r}_{12}=(0,0,1) \)
\( \vec{r}_{22}=(0,0,0) \)
Here, we used the suffix 1 and 2 for derivatives with respect to u and v and respectively, and similarly for higher derivatives.
Now, first order fundamental coefficients are
\( E=\vec{r}_1^2=(1,0,v)^2=1+v^2 \)
\( F=\vec{r}_1. \vec{r}_2=(1,0,v).(0,1,u)=uv \)
\( G=\vec{r}_2^2=(1,0,u)^2=1+u^2 \)
Next,we have to compute second fundamental cofficients,for this
\( H\vec{N}=\vec{r}_1\times \vec{r}_2 \)
or \( H\vec{N}=(1,0,v)\times (0,1,u) \)
or \( H\vec{N}=(-v,-u,1) \) (A)
Taking magnitude, we get
\( H=\sqrt{1+u^2+v^2}\)
And substituting H in (A) we get
\( \vec{N}=\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}\)
Hence, the second order fundamental coefficients are
\(L= \vec{r}_{11}.\vec{N}=(0,0,0).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=0\)
\(M= \vec{r}_{12}.\vec{N}=(0,0,1).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=\frac{1}{\sqrt{1+u^2+v^2}}\)
\(N= \vec{r}_{22}.\vec{N}=(0,0,0).\frac{(-v,-u,1)}{\sqrt{1+u^2+v^2}}=0\)
This completes the solution

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Find fundamental coefficients of following surface

1. Monge’s form: \( \vec{r}=( x,y,f( x,y ) )\)
   

   Answer:
   \( \vec{N}=\left\{-\frac{f_1}{\sqrt{1 + f_1^2 + f_2^2}}, -\frac{f_2}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{1}{\sqrt{1 + f_1^2 + f_2^2}}\right\} \)
   Cofficients: \( \left\{1 + f_1^2, f_1 f_2, 1 + f_2^2, \frac{f_{11}}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{f_{12}}{\sqrt{1 + f_1^2 + f_2^2}}, \frac{f_{22}}{\sqrt{1 + f_1^2 + f_2^2}}\right\} \)

2. Surface of revolution: \(\vec{r}=( u \cos v,u \sin v,f( u ) )\)
   

   Answer:
   \( \vec{N}=\left\{-\frac{f_1 u \cos v}{H}, -\frac{f_1 u \sin v}{H}, \frac{u}{H}\right\} \)
   Cofficients: \( \left\{1 + f_1^2, 0, u^2, \frac{f_{11} u}{H}, 0, \frac{f_1 u^2}{H}\right\} \)

3. Conoidal surface:\( \vec{r}=( u \cos v,u \sin v,f( v ) ) \)
   

   Answer:
   \( \vec{N}=\left\{\frac{f_2 \sin v}{H}, -\frac{f_2 \cos v}{H}, \frac{u}{H}\right\}\)
   Cofficients: \( \left\{1, 0, f_2^2 + u^2, 0, -\frac{f_2}{H}, \frac{f_{22} u}{H}\right\} \)

4. Right helicoid: \( \vec{r}=( u \cos v,u\sin v,cv ) \)
   

   Answer:
   \( \vec{N}=\left\{\frac{c \sin v}{H}, -\frac{c \cos v}{H}, \frac{u}{H}\right\}\)
   Cofficients: \( \left\{1, 0, c^2 + u^2, 0, -\frac{c}{H}, 0\right\} \)

5. Plane surface: \( \vec{r}=( u,v,u+v) \)
   

   Answer:
   \( \vec{N}=\left\{\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right\}\)
   Cofficients: \( \{3, 1, 1, 0, 0, 0\} \)

6. Saddle surface: \( \vec{r}=( u,v,uv) \)
   

   Answer:
   \( \vec{N}=\left\{-\frac{v}{\sqrt{1 + u^2 + v^2}}, -\frac{u}{\sqrt{1 + u^2 + v^2}}, \frac{1}{\sqrt{1 + u^2 + v^2}}\right\}\)
   Cofficients: \( \left\{1 + v^2, uv, 1 + u^2, 0, \frac{1}{\sqrt{1 + u^2 + v^2}}, 0\right\}\)

7. Saddle surface: \( \vec{r}=( u+v,u-v,uv ) \)
   

   Answer:
   \( \vec{N}=\left\{\frac{u + v}{\sqrt{2} \sqrt{2 + u^2 + v^2}}, \frac{-u + v}{\sqrt{2} \sqrt{2 + u^2 + v^2}}, -\frac{\sqrt{2}}{\sqrt{2 + u^2 + v^2}}\right\} \)
   Cofficients: \( \left\{2 + v^2, uv, 2 + u^2, 0, -\frac{\sqrt{2}}{\sqrt{2 + u^2 + v^2}}, 0\right\}\)

8. Paraboloid:\(\vec{r}=( u,v,u^2+v^2 ) \)
   

   Answer:
   \( \vec{N}=\left\{\frac{v}{\sqrt{2}}, -\frac{v}{\sqrt{2}}, 0\right\}\)
   Cofficients: \( \{a^2, 0, 1, -a, 0, 0\} \)

9. Cylinder: \( \vec{r}=(a\cos u, a\sin u,v ) \)
   

   Answer:
   \( \vec{N}=\{\cos u, \sin u, 0\}\)
   Cofficients: \( \{2 + 4 u^2, 4 u v, 4 v^2, 0, 0, 0\} \)

10. Cone: \( \vec{r}=(v\cos u, v\sin u,v ) \)
    

    Answer:
    \( \vec{N}=\left\{\frac{v \cos u}{\sqrt{2}}, \frac{v \sin u}{\sqrt{2}}, -\frac{v}{\sqrt{2}}\right\} \)
    Cofficients: \( \left\{v^2, 0, 2, -\frac{v^2}{\sqrt{2}}, 0, 0\right\} \)

11. Sphere: \( \vec{r}=(\sin u \cos v,\sin u \sin v, \cos u) \)
    

    Answer:
    \( \vec{N}=\left\{\frac{\cos v \sin^2 u}{H}, \frac{\sin^2 u \sin v}{H}, \frac{\cos u \sin u}{H}\right\} \)
    Cofficients: \( \left\{1, 0, \sin^2 u, -\frac{\sin u}{H}, 0, -\frac{\sin^3 u}{H}\right\} \)

12. Hyperboloid: \( 2z=7x^2+6xy-y^2 \)at origin
    

    Answer:
    \( \vec{N}= \{0,0,1\}\)
    Cofficients: \( \left\{ 1,0,1,7,3,-1\right\} \)

13. Minimal surface: \( e^z \cos x=\cos y \)
    

    Answer:
    \( \vec{N}= \left\{-\frac{\tan x}{H}, \frac{\tan y}{H}, \frac{1}{H}\right\} \)
    Cofficients: \( \left\{\sec^2 x, -\tan x \tan y, \sec^2 y, \frac{\sec^2 x}{H}, 0, -\frac{\sec^2 y}{H}\right\} \)

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Fundamental Forms of Surface

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We recall that, space curve is uniquely determined by two local invariant quantities: curvature and torsion.Similarly, surface is uniquely determined by two local invariant quantities: first and second fundamental form of surface

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First Fundamental Form of Surface

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Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surface, then, quadratic differential form in \( du,dv \) given by
\( I:Edu^2+2Fdudv+Gdv^2 \) where \( E=\vec{r_1}^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)
is called first fundamental form of the surface. The coefficients \( E,F,G\) are called the first fundamental coefficients or magnitude of first order.

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Geometry of First Fundamental Form of Surface

Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surfacein which \( P \) and \( Q \) are two neighboring points with
\( \overrightarrow{OP}=\vec{r} \) and \( \overrightarrow{OQ}=\vec{r}+d\vec{r}\)
Then
\( \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP} \)
or \( d\vec{r}=\vec{r}( u+du,v+dv )-\vec{r}( u,v) \)
or \( d\vec{r}=\vec{r}( u,v )+( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}(\vec{r}_1du+\vec{r}_2dv )^2+...-\vec{r}( u,v ) \)
or \( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac1{2!}{{( \vec{r}_1du+\vec{r}_2dv )}^2}+\ldots \)
The first order approximation is
\(dr=( \vec{r}_1du+\vec{r}_2dv ) \)
Squaring on both sides, we get
\( ds^2=(\vec{r}_1du+\vec{r}_2dv )^2 \)
or \( ds^2=\vec{r}_1^2du^2+2\vec{r}_1\vec{r}_2dudv+\vec{r}_2^2dv^2 \)
or \( ds^2=Edu^2+2Fdudv+Gdv^2\) where \( E=\vec{r}_1^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)

Note

1. For \( u\) curve \( ds^2=Edu^2 \)
2. For \( v \) curve \( ds^2=Gdv^2 \)

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By the ordinary property of vector operation,
\( ( \vec{r}_1\times \vec{r}_2)^2=\vec{r}_1^2. \vec{r}_2^2-( \vec{r}_1.\vec{r}_2 )^2 \)
or \( (H \vec{N})^2=E.G-F^2 \)
or \(H^2=E.G-F^2 \), which is always positive.

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Property 1 : First Fundamental Form is Positive

The first fundamental form of the surface is positive definite in \( du \) and \( dv \).
Proof
Let \(S:\vec{r}=\vec{r}( u,v ) \) be a surface, then
\( H^2=E.G-F^2 \) is always positive
Now, assume that \( E >0 \), then
\(I=Edu^2+2Fdudv+Gdv^2 \)
or \(I=\frac{1}{E}( E^2du^2+2EFdudv+EGdv^2 )\)
or \(I=\frac{1}{E}[ ( Edu+Fdv )^2+( EG-F^2 )dv^2 ]\)
or \(I=\frac{1}{E}[ ( Edu+Fdv )^2+H^2dv^2 ]\)
Here, \(I \) has two possibilities, one is \( I=0 \) and other is \( I>0 \)
If possible, suppose that
\( I=0 \)
Then, we must have
\( \frac{1}{E}[ ( Edu+Fdv )^2+H^2dv^2 ]=0 \)
or \( ( Edu+Fdv )^2+H^2dv^2=0 \)
or \( ( Edu+Fdv )^2=0 \) and \( H^2 dv^2=0 \)
or \( ( Edu+Fdv )^2=0 \) and \( dv^2=0 \)
or \(Edu+Fdv=0 \) and \( dv=0 \)
or \(Edu =0 \) and \( dv=0 \)
or \( du=0 \) and \( dv= 0 \) , which is not possible.
Hence, first fundamental form is positive definite in \( du \) and \( dv \).

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Property 2 : First Fundamental Form is Invariant

The first fundamental form is invariant under parametric transformation.
Proof
Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surface, and parameters \( ( u,v ) \) is transformed into another set \(( U,V ) \) set of parameters such that
\(U =\phi ( u,v ) \) and \( V=\psi ( u,v ) \)
Now
\(I = Edu^2+2Fdudv+Gdv^2 \) where \( E=\vec{r}_1^2,F=\vec{r}_1.\vec{r}_2,G=\vec{r}_2^2 \)
or \( I =(\textcolor{red}{ \vec{r}_1} du+ \textcolor{red}{ \vec{r}_2} dv )^2 \)
or \( I =( \textcolor{red}{ \frac{\partial \vec{r}} {\partial u}} du + \textcolor{red}{ \frac{\partial \vec{r}} {\partial v}} dv)^2 \)
or \(I = ( [ \textcolor{red}{ \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} } ] du + [ \textcolor{red}{ \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} }] dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} du + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} du + \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} dv + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial u} du +\frac{\partial \vec{r}} {\partial U} \frac{\partial U} {\partial v} dv + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial u} du + \frac{\partial \vec{r}} {\partial V} \frac{\partial V} {\partial v} dv)^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U} [\frac{\partial U} {\partial u} du + \frac{\partial U} {\partial v} dv] + \frac{\partial \vec{r}} {\partial V} [ \frac{\partial V} {\partial u} du + \frac{\partial V} {\partial v} dv])^2 \)
or \(I = ( \frac{\partial \vec{r}} {\partial U}dU + \frac{\partial \vec{r}} {\partial V}dV)^2 \)
or \(I = EdU^2+2FdU dV+GdV^2 \) where \( E= (\frac{\partial \vec{r}} {\partial U})^2,F=\frac{\partial \vec{r}} {\partial U}.\frac{\partial \vec{r}} {\partial V},G= (\frac{\partial \vec{r}} {\partial V})^2 \)
This shows that, first fundamental form is invariant under parametric transformation.

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Elements of surface area

Let \(S:\vec{r}=\vec{r}( u,v )\) be a surface. If \(\Delta R=PQRS\) is a small region on the surface with
\(P( u,v ),Q( u+du,v ),R( u+du,v+dv )\) and \(( u,v+dv ) \).
Then, \( PQRS \) tends to a parallelogram when \( du, dv \) are very small and positive
Thus
Area of \( PQRS =| \vec{PQ} \times \vec{PS} | \) (i)
Here
\( \vec{PQ}=\vec{OQ}-\vec{OP} \)
or \( \vec{PQ}=\vec{r}( u+du,v )-\vec{r}( u,v ) \)
or \( \vec{PQ}=\vec{r}( u,v )+\vec{r}_1du+... -\vec{r}( u,v ) \)
Taking first order approximation, we get
\( \vec{PQ}=\vec{r}_1du \) (ii)
Similarly we get
\( \vec{PS}=\vec{r}_2dv \) (iii)
Thus, from (i), (ii) and (iii), we have
Area of \( PQRS = | \vec{r}_1 du \times \vec{r}_2dv | \)
or Area of \( PQRS =| \vec{r}_1 du \times \vec{r}_2 dv | \)
or Area of \( PQRS =| \vec{r}_1 \times \vec{r}_2 | du dv\)
or Area of \( PQRS =H du dv\)

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Second Fundamental Form of the Surface

Let \(S:\vec{r}=\vec{r}( u,v )\) be a surface.Then quadratic differential form in \( du, dv \)
\( II:Ldu^2+2Mdudv+Ndv^2 \) where \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},N=\vec{r}_{22}.\vec{N} \)
is called second fundamental form of the surface.
The coefficients \(L,M,N \) are called the second fundamental coefficients or magnitude of second order.

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Geometry of Second Fundamental Form of the Surface

Length of perpendicular on the tangent plane from neighborhood point on the surface is
\(\frac12( Ldu^2+2Mdudv+Ndv^2 ) \)
Proof
Let \( S:\vec{r}=\vec{r}( u,v ) \) be a surfacein which \( P \) and \( Q \) are two neighboring points with
\( \vec{OP}=\vec{r} \) and \( \vec{OQ}=\vec{r}+d\vec{r}\)
Then
\( \vec{PQ}=\vec{OQ}-\vec{OP} \)
or \( d\vec{r}=\vec{r}( u+du,v+dv )-\vec{r}( u,v) \)
or \( d\vec{r}=\vec{r}( u,v )+( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2+...-\vec{r}( u,v ) \)
or \( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2+... \)
The second order approximation is
\( d\vec{r}=( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_1du+\vec{r}_2dv )^2 \)
or \( d\vec{r}= (\vec{r}_1du+\vec{r}_2dv )+ \frac{1}{2} ( \vec{r}_{11} du^2 +2 \vec{r}_{12} du dv+ \vec{r}_{22} dv^2) \)
Now, let M be the projection of Q on the tangent plane at P, then
\( QM=\) Projection of \( \vec{PQ} \) on the normal at P
or \( QM=\vec{PQ}.\vec{N}\)
or \( QM=[ ( \vec{r}_1du+\vec{r}_2dv )+\frac{1}{2!}( \vec{r}_{11}du^2+2\vec{r}_{12}dudv+\vec{r}_{22}dv^2 ) ].\vec{N} \)
or \( QM=\frac{1}{2!}( \vec{r}_{11}.\vec{N}du^2+2\vec{r}_{12}.\vec{N}dudv+\vec{r}_{22}.\vec{N}dv^2 )\)
or \( QM=\frac{1}{2}( Ldu^2+2Mdudv+Ndv^2 )\) where \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},G=\vec{r}_{22}.\vec{N} \)

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Alternative forms of \( L, M, N \)

As we know that \(\vec{r_1}\) is tangent to u-curve; and \(\vec{r_2}\) is tangent to v-curve, so \(\vec{r_1}\) and \(\vec{r_2}\) are tangents of the surface \( S:\vec{r}=\vec{r}( u,v ) \). Therefore, \(\vec{r_1}\) and \(\vec{r_2}\) ate perpendicular to the normal vector \(\vec{N}\). Also, we know that \(L=\vec{r}_{11}.\vec{N},M=\vec{r}_{12}.\vec{N},G=\vec{r}_{22}.\vec{N} \). Now, the fundamental coefficients \(L,M,N\) can be alternatively explained as below.

1. Also, we know that
   \(\vec{r}_1. \vec{N}=0 \)(1)
   Differentiating (1) w. r. to u, we, get
   \(\vec{r}_{11}. \vec{N}+\vec{r}_1. \vec{N}_1=0 \)
   or \(L+\vec{r}_1. \vec{N}_1=0 \)
   or \(\vec{r}_1. \vec{N}_1=-L \)

2. Also, we know that
   \(\vec{r}_1. \vec{N}=0 \)(1)
   Differentiating (1) w. r. to v, we, get
   \(\vec{r}_{12}. \vec{N}+\vec{r}_1. \vec{N}_2=0 \)
   or \(M+\vec{r}_1. \vec{N}_2=0 \)
   or \(\vec{r}_1. \vec{N}_2=-M \)

3. Also, we know that
   \(\vec{r}_2. \vec{N}=0 \)(1)
   Differentiating (1) w. r. to u, we, get
   \(\vec{r}_{21}. \vec{N}+\vec{r}_2. \vec{N}_1=0 \)
   or \(M+\vec{r}_2. \vec{N}_1=0 \)
   or \(\vec{r}_2. \vec{N}_1=-M \)

4. Also, we know that
   \(\vec{r}_2. \vec{N}=0 \)(1)
   Differentiating (1) w. r. to v, we, get
   \(\vec{r}_{22}. \vec{N}+\vec{r}_2. \vec{N}_2=0 \)
   or \(N+\vec{r}_2. \vec{N}_2=0 \)
   or \(\vec{r}_2. \vec{N}_2=-N \)

Based on the cofficients of L, M and N, we can write that

1. \([\vec{r}_1,\vec{r}_2,\vec{r}_{11}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{11}=H \vec{N} .\vec{r}_{11}=HL\)
2. \([\vec{r}_1,\vec{r}_2,\vec{r}_{12}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{12}=H \vec{N} .\vec{r}_{12}=HM\)
3. \([\vec{r}_1,\vec{r}_2,\vec{r}_{21}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{21}=H \vec{N} .\vec{r}_{21}=HM\)
4. \([\vec{r}_1,\vec{r}_2,\vec{r}_{22}]=\vec{r}_1 \times \vec{r}_2 .\vec{r}_{22}=H \vec{N} .\vec{r}_{22}=HN\)

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Nature of points on a surface

The second fundamental form of a surface measures osculation paraboloid, which helps to determines nature of points on the surface. These points are as follows

1. Case 1: Parabolic
   
   A point on a surface is called parabolic point if
   \(LN-M^2=0;L^2+M^2+N^2 \ne 0\)
   In parabolic point, there exists a line in tangent plane whose normal curvature is zero. So, at this point, exactly one of principal curvatures \(K_1\) and \(K_2\) is zero. Since one normal curvature is zero, the direction corresponding to the zero principal curvature will be the direction of the asymptotic curve. The nbd point of the surface lies on same side of the tangent plane. An example of such point is shown in a paraboloid cylinder in figure.
2. Case 2: Hyperbolic
   
   A point is called hyperbolic point if
   \(LN-M^2<0\)
   In hyperbolic point, there exists two lines in tangent plane whose normal curvatures: \(K_1\) and \(K_2\) have opposite sign. In such point, there exists two asymptotic direction. The nbd point of the surface lies on both sides of the tangent plane. An example of such point is shown in a hyperboloid surface in figure
3. Case 3: Elliptic
   
   A point is called elliptic point if
   \(LN-M^2>0\)
   In elliptic point, there exists tangent plane whose normal curvatures: \(K_1\) and \(K_2\) have same sign. In such point, there exists no asymptotic direction. The nbd point of the lies on same side of the tangent plane. An example of such point is shown in an elliptic surface in figure
4. Case 4: Planner
   
   A point is called planner point if
   \(LN-M^2=0;L^2+M^2+N^2= 0\)
   If \(L,M,N\) and \(LN-M^2\) are all zero, then surface is planar, means both of \(K_1\) and \(K_2\) is zero.

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Equations of tangent and normal to a parabola

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Tangent to Parabola at \( (x_1, y_1) \)

Answer 👉 Click Here

Tangent to Parabola \(y^2=4ax\) at \( (x_1, y_1) \)

Let the point of tangency be \( (x_1, y_1) \) on the parabola \(y^2=4ax\), we can write that
\( y_1^2=4ax_1 \) (i)
Now the equation of tangent to the parabola at \( (x_1, y_1) \) is
\( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is
\( y^2 = 4ax \)
Differentiate both sides with respect to \( x \), we get
\( 2y \frac{dy}{dx} = 4a \)
or \( \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
Now, the equation of the tangent line at a point \( (x_1, y_1) \) with slope \( m \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{2a}{y_1}(x - x_1) \) Answer
or \( y y_1 - y_1^2 = 2a(x - x_1) \)
or \( y y_1 - 4ax_1 = 2ax - 2ax_1 \) From (i): \( y_1^2=4ax_1 \)
or \( y y_1 = 2ax + 2ax_1 \)
or \( y y_1 = 2a(x + x_1) \)
NOTE
Now, the equation of normal is
\( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)

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Tangent to Parabola \(x^2=4ay\) at \( (x_1, y_1) \)

Let the point of tangency be \( (x_1, y_1) \) on the parabola \( x^2 = 4ay \), we can write that
\( x_1^2 = 4ay_1 \) (i)
Now the equation of tangent to the parabola at \( (x_1, y_1) \) is
\( y - y_1 = m(x - x_1) \) (A)
The slope of the tangent to the parabola is
\( x^2 = 4ay \)
Differentiate both sides with respect to \( x \), we get
\( 4a \frac{dy}{dx} =2x \)
or \( \frac{dx}{dy} = \frac{2x}{4a} = \frac{x}{2a} \)
Now, the equation of the tangent line at a point \( (x_1, y_1) \) with slope \( m \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{x_1}{2a}(x - x_1) \) Answer
or \( 2a(y - y_1) = xx_1 - x_1^2) \)
or \( 2ay - 2ay_1 = xx_1 - 4ay_1) \) From (i): \( x_1^2 = 4ay_1 \)
or \( 2ay - 2ay_1 = xx_1 - 4ay_1) \)
or \( x x_1 = 2a(y + y_1) \)
NOTE
Now, the equation of normal is
\( y - y_1 = -\frac{2a}{x_1}(x - x_1) \)

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Condition of Tangenncy to Parabola

Answer 👉 Click Here

Condition of Tangentcy: line \( y = mx + c \) to the parabola \( y^2 = 4ax \)

Let \( y = mx + c \) be a tangent to the parabola \( y^2 = 4ax \), then they must intersect at exactly one point.
Substituting \( y = mx + c \) on \( y^2 = 4ax \), we get
\( (mx + c)^2 = 4ax \)
\( m^2x^2 + 2mcx + c^2 = 4ax \)
\( m^2x^2 + (2mc - 4a)x + c^2 = 0 \)
This is a quadratic equation in \( x \), this quadratic equation must have exactly one solution, so the discriminant must be zero, thus we have
\( b^2 =4ac\)
or\( (2mc - 4a)^2 = 4m^2c^2 \)
or\( 2mc - 4a = \pm 2mc \) (A) Taking positive sign from (A), we get
\( 2mc - 4a = 2mc \)
or \( -4a = 0 \)
or \( a = 0 \) (not possible as \( a \neq 0 \) for a parabola)
Taking negative sign from (A), we get
\( 2mc - 4a = -2mc \)
or\( 4mc = 4a \)
or \( mc = a \) Therefore, the condition for the line \( y = mx + c \) to be tangent to the parabola \( y^2 = 4ax \) is
\( mc = a \) or \(c=\frac{a}{m} \)

NOTE

1. Equation of tangent:
   The equation of tangent line to the parabola \( y^2 = 4ax \) is
   \( y = mx + c \)
   or\( y = mx + \frac{a}{m} \) Answer
2. The point of contact:
   The point of contact of tangent \( y = mx + \frac{a}{m} \) to the parabola \( y^2 = 4ax \) is
   \( (x_1,y_1) \)
   Then
   \( y = mx + \frac{a}{m} \) (A)
   \( yy_1 = 2a(x + x_1) \) (B)
   must be identical
   So, we can write that
   \( \frac{1}{y_1}= \frac{m}{2a}=\frac{a/m}{2ax_1} \)
   Equationg first and second part gives
   \( \frac{1}{y_1}= \frac{m}{2a} \) or \(y_1=\frac{2a}{m}\)
   Equationg second and third part gives
   \( \frac{m}{2a}=\frac{a/m}{2ax_1} \) or \(x_1=\frac{a}{m^2}\)
   Hence, the point of contact is
   \( (x_1,y_1)=(\frac{a}{m^2},\frac{2a}{m}) \) Answer

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Condition of Tangentcy: line \( y = mx + c \) to the parabola \( x^2 = 4ay \)

Let \( y = mx + c \) be a tangent to the parabola \( x^2 = 4ay \), then they must intersect at exactly one point.
Substituting \( y = mx + c \) into \( x^2 = 4ay \), we get
\( x^2 = 4a(mx + c) \)
\( x^2 = 4amx + 4ac \)
\( x^2 - 4amx - 4ac = 0 \)
This is a quadratic equation in \( x \), which must have exactly one solution, so the discriminant must be zero, thus we have
\( b^2 = 4ac \)
or\( (4am)^2 - 4(1)(-4ac) = 0 \)
or\( 16a^2m^2 + 16ac = 0 \)
or\( 16a(am^2 + c) = 0 \)
Since \( a \neq 0 \) for a parabola, we have:
\( am^2 + c = 0 \)
or\( c = -am^2 \)
Therefore, the condition for the line \( y = mx + c \) to be tangent to the parabola \( x^2 = 4ay \) is
\( c = -am^2 \)

NOTE

1. Equation of tangent:
   The equation of tangent line to the parabola \( x^2 = 4ay \) is
   \( y = mx + c \)
   or\( y = mx -am^2 \)
2. The point of contact:
   The point of contact of tangent \( y = mx -am^2 \) to the parabola \( x^2 = 4ay \) is
   \( (x_1,y_1) \)
   Then
   \( y = mx -am^2 \) (A)
   \( x x_1 = 2a(y + y_1) \) (B)
   must be identical
   So, we can write that
   \( \frac{1}{2a}= \frac{m}{x_1}=\frac{am^2}{2ay_1} \)
   Equationg first and second part gives
   \( \frac{1}{2a}= \frac{m}{x_1} \) or \(x_1=\frac{m}{2a}\)
   Equationg first and third part gives
   \(\frac{1}{2a}=\frac{am^2}{2ay_1} \) or \(y_1=am^2\)
   Hence, the point of contact is
   \( (x_1,y_1)=(\frac{m}{2a},am^2) \)

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Normal to Parabola at \( (x_1, y_1) \)

Answer 👉 Click Here

Normal to Parabola \(y^2=4ax\) at \( (x_1, y_1) \)

At \( (x_1, y_1) \), the equation of normal is
\( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is
\( y^2 = 4ax \)
or \( 2y \frac{dy}{dx} = 4a \)
or \( \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
The slope of the normal to the parabola \( y^2 = 4ax \) at \( (x_1, y_1) \) is
\( -\frac{y_1}{2a} \)
Now, the equation of the normal line at a point \( (x_1, y_1) \) to the parabola \( y^2 = 4ax \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)

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Normal to Parabola \(y^2=4ax\) in Slope form

The equation of the normal to the parabola \( y^2 = 4ax \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
where
\(m = -\frac{y_1}{2a} \)
So, we have
\(y_1=-2am \) and \(x_1=\frac{y_1^2}{4a}=\frac{(-2am)^2}{4a}= am^2 \)
Hence the equation of normal is
\( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
or \( y - (-2am ) = m[x - (am^2)] \)
or \( y =mx-2am -am^3 \)

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Normal to Parabola \(x^2=4ay\) at \( (x_1, y_1) \)

At \( (x_1, y_1) \), the equation of normal is
\( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is
\( x^2 = 4ay \)
Differentiate both sides with respect to \( x \), we get
\( 4a \frac{dy}{dx} =2x \)
or \( \frac{dx}{dy} = \frac{2x}{4a} = \frac{x}{2a} \)
The slope of the normal to the parabola \( x^2 = 4ay \) at \( (x_1, y_1) \) is
\( -\frac{2a}{x_1} \)
Now, the equation of the normal line at a point \( (x_1, y_1) \) to the parabola \( x^2 = 4ay \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{2a}{x_1} (x - x_1) \)

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Normal to Parabola \(x^2=4ay\) in Slope form

The equation of the normal to the parabola \( x^2 = 4ay \) is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{2a}{x_1}(x - x_1) \)
where
\(m =-\frac{2a}{x_1} \)
So, we have
\(x_1=-\frac{2a}{m} \) and \(y_1=\frac{x_1^2}{4a}=\frac{4a^2}{m^24a}= \frac{a}{m} \)
Hence the equation of normal is
\( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
or \( y - (-\frac{2a}{m} ) = m[x - (\frac{a}{m} )] \)
or \( y =mx-\frac{2a}{m}-a \)

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Tangent and Normal to Parabola in Parametric Form

Answer 👉 Click Here

Tangent and Normal to Parabola \(y^2=4ax\) in parametric form

Let \(y^2=4ax\) be a parabola in which parametric coordinate is \((at^2,2at)\)
\( y_1^2=4ax_1 \) (i)
Now the equation of tangent to the parabola is
\( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is
\( m= \frac{2a}{y} =\frac{2a}{2at}=\frac{1}{t}\)
Now, the equation of the tangent is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{1}{t}(x - x_1) \)
Now, the equation of normal is
\( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -t(x - x_1) \)

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Exercise

1. Find the equation of tangent and normal to the parabola
   1. \( y^2=8x\) at (2,-4)
      

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( y^2 = 4ax \), so we have \( 4a = 8 \), so \( a = 2 \)
         Also, the slope of tangent to the parabola at \( (x_1, y_1)=(2,-4) \) is
         \( m_1 = \frac{2a}{y_1} =\frac{2.2}{-4}=-1\)
         Now, equation of tangent to the parabola is
         \( y-y_1=m_1(x-x_1)\)
         \( y+4 = (-1) (x - 2) \)
         or\( y+4=-x+2 \)
         or\( x + y + 2 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, slope of normal to the parabola at \( (x_1, y_1)=(2,-4) \) is
         \( m_2 = \frac{y_1}{2a} =1\)
         Now, equation of normal to the parabola is
         \( y-y_1=m_2(x-x_1)\)
         \( y+4 = (1) (x - 2) \)
         or\( y+4=x-2 \)
         or\( x - y -6= 0 \) Answer: normal line
   2. \( x^2=12y\) at (-6,3)

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( x^2 = 4ay \), so we have \( 4a = 12 \), so \( a = 3 \)
         Also, the slope of tangent to the parabola at \( (x_1, y_1) = (-6, 3) \) is
         \( m_1 = \frac{x_1}{2a} =\frac{-6}{2 \cdot 3} = -1 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 3 = (-1) (x + 6) \)
         or\( y - 3 = -x - 6 \)
         or\( x + y + 3 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, the slope of the normal to the parabola at \( (x_1, y_1) = (-6, 3) \) is
         \( m_2 = -\frac{1}{m_1} = 1 \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 3 = (1) (x + 6) \)
         or\( y - 3 = x + 6 \)
         or\( x - y + 9 = 0 \) Answer: normal line
   3. \( x^2=4y\) at the point whose abscissa is 6

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( x^2 = 4y \), so we have \( 4a = 4 \), so \( a = 1 \)
         Given \( x_1 = 6 \), we find the corresponding \( y_1 \) by substituting into the equation:
         \( y_1 = \frac{x_1^2}{4} = \frac{6^2}{4} = 9 \)
         So, the point of tangency is \( (6, 9) \).
         Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (6, 9) \) is
         \( m_1 = \frac{x_1}{2a} = \frac{6}{2 \cdot 1} = \frac{6}{2} =3 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 9 = 3(x - 6) \)
         or\( y - 9 = 3x - 18 \)
         or\( 3x - y -9 = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         Now, the slope of the normal to the parabola at \( (x_1, y_1) = (6, 9) \) is
         \( m_2 = -\frac{1}{m_1} = -\frac{1}{3} \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 9 = -\frac{1}{3}(x - 6) \)
         or\( 3y - 27 = -x + 6 \)
         or\( x +3 y - 33 = 0 \) Answer: normal line
   4. \( y^2=16ax\) at the point whose ordinate is -4a

      Answer 👉 Click Here

      1. Equation of the Tangent
         The parabola is \( y^2 = 16ax \).
         Given \( y_1 = -4a \), we find the corresponding \( x_1 \) by substituting into the equation:
         \( (-4a)^2 = 16a \cdot x_1 \implies 16a^2 = 16a \cdot x_1 \implies x_1 = a \)
         So, the point of tangency is \( (a, -4a) \).
         Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (a, -4a) \) is
         \( m_1 = \frac{8a}{y_1} = \frac{8a}{-4a} = -2 \)
         Now, the equation of the tangent to the parabola is
         \( y - y_1 = m_1(x - x_1) \)
         \( y + 4a = -2(x - a) \)
         or\( y + 4a = -2x + 2a \)
         or\( 2x + y + 2a = 0 \) Answer: tangent line
         
      2. Equation of the Normal
         The slope of the normal to the parabola at \( (x_1, y_1) = (a, -4a) \) is
         \( m_2 = \frac{-1}{m_1} = \frac{1}{2} \)
         Now, the equation of the normal to the parabola is
         \( y - y_1 = m_2(x - x_1) \)
         \( y + 4a = \frac{1}{2}(x - a) \)
         or\( 2(y + 4a) = x - a \)
         or\( x - 2y - 9a = 0 \) Answer: normal line
   5. \( y^2=12x\) at each end of the latus rectum

      Answer 👉 Click Here

      We know that, end of the latus rectum are \(a,\pm 2a)\).
      For the parabola \( y^2 = 12x \), we have \( 4a = 12 \), so \( a = 3 \). The end of latus rectum are \( (3, 6) \) and \( (3, -6) \).
      

      1. At the End \( (3, 6) \)
         Equation of the Tangent
         The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, 6) \) is
         \( m_1 = \frac{y_1}{2a} = \frac{6}{6} = 1 \)
         Now, the equation of the tangent is
         \( y - y_1 = m_1(x - x_1) \)
         \( y - 6 = 1(x - 3) \)
         or\( y - 6 = x-3 \)
         or\( x-y+3=0 \) Answer: tangent line
         Equation of the Normal
         The slope of the normal is the negative reciprocal of the tangent’s slope:
         \( m_2 = -\frac{1}{m_1} = -1 \)
         Now, the equation of the normal is
         \( y - y_1 = m_2(x - x_1) \)
         \( y - 6 = -1(x - 3) \)
         or\( y - 6 = -x + 3 \)
         or\( x + y - 9 = 0 \) Answer: normal line
      2. At the End \( (3, -6) \)
         
         
         The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, -6) \) is
         \( m_1 = \frac{y_1}{2a} = \frac{6}{-6} = -1 \)
         Now, the equation of the tangent is
         \( y - y_1 = m_1(x - x_1) \)
         \( y + 6 = -1(x - 3) \)
         or\( y + 6 = -x+3 \)
         or\( x+y+3=0 \) Answer: tangent line
         Equation of the Normal
         The slope of the normal is the negative reciprocal of the tangent’s slope:
         \( m_2 = -\frac{1}{m_1} = 1 \)
         Now, the equation of the normal is
         \( y - y_1 = m_2(x - x_1) \)
         \( y + 6 = 1(x - 3) \)
         or\( y + 6 = x - 3 \)
         or\( x-y -9 = 0 \) Answer: normal line
2. Solve the followings
   1. Prove that the line \(3x+4y+6=0\) is tangent to the parabola \(2y^2=9x\) and find its point of contact
      

      Answer 👉 Click Here

      
      The given parabola is \(2y^2 = 9x\), which is
      \( y^2 = \frac{9}{2} x \) (i)
      Thus,
      \( a = \frac{9}{8} \)
      The given line is
      \(3x + 4y + 6 = 0\)
      or \( y = -\frac{3}{4}x - \frac{3}{2} \) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(- \frac{3}{2} =\frac{\frac{9}{8}}{ -\frac{3}{4}}\) TRUE
      So, the line (ii) is tangent to the parabola (i)
      Next, to find the point of contact, we substitute \(x = -\frac{4y + 6}{3}\) into the parabola , so that
      \( 2y^2 = 9 \left(-\frac{4y + 6}{3}\right) \)
      \( 2y^2 = -12y - 18 \)
      \( 2y^2 + 12y + 18 = 0 \)
      \( y = -3 \)
      Substitute \( y = -3 \) into the line equation \(3x + 4y + 6 = 0\) , we get
      \( 3x + 4(-3) + 6 = 0 \)
      \( x=2 \)
      So, the point of contact is \( (2, -3) \)
   2. Prove that the line \(lx+my+n=0\) touches the parabola \(y^2=4ax\) if \(ln =am^2\)
      

      Answer 👉 Click Here

      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(lx+my+n=0\)
      or \( y = -\frac{l}{m}x - \frac{n}{m} \) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(- \frac{n}{m} =\frac{a}{ -\frac{l}{m}}\) TRUE
      \(ln =am^2\)
   3. For what value of a, will the straight line \(y=2x+3\) touch the parabola \(y^2=4ax\)
      

      Answer 👉 Click Here

      
      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(y=2x+3\) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(3 =\frac{a}{2}\) TRUE
      \(a=6\)
   4. If the line \(2x+4y=3\) touches the parabola \(y^2=4ax\) find the length of the latus rectum
      

      Answer 👉 Click Here

      
      The given parabola is
      \(y^2=4ax\) (i)
      The given line is
      \(2x+4y=3\)
      \(y=-\frac{1}{2}x+\frac{3}{4}\) (ii)
      The condition for line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(\frac{3}{4} =\frac{a}{-\frac{1}{2}}\) TRUE
      \(a=-\frac{3}{8}\)
      Now, the length of rectum is
      \(4a=4 \times \frac{3}{8}=\frac{3}{2} \)
3. Solve the followings
   1. Find the equation of tangent to the parabola \(y^2=4x\)
      (i) parallel to the line \(x-2y+6=0\).
      (ii) perpendicular to the line \(2x-y=4\)
      Also find the point of contact.

      Answer 👉 Click Here

      (i) parallel to the line \(x-2y+6=0\). Also find the point of contact.
      

      
      The given parabola is
      \(y^2=4x\) (i)
      Thus,
      \( 4a = 4\) gives \(a=1\)
      The equation of tangent to parabola, which is parallel to \( x - 2y + 6 = 0 \) is
      \( x - 2y + k = 0 \) (A)
      \( 2y = x + k \)
      \( y = \frac{1}{2}x +\frac{k}{2} \) (ii)
      The condition for the line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(\frac{k}{2} =\frac{1}{\frac{1}{2}}\)
      \(k=4\)
      Hence, the required equation of tangent line is, from (A)
      \( x - 2y + 4 = 0 \)
      Next, to find the point of contact, we substitute \(x = 2y - 4\) into the parabola , so that
      \( y^2 = 4 \left(2y - 4 \right) \)
      \( y^2 = 8y - 16 \)
      \( y^2 - 8y +16 = 0 \)
      \( y = 4 \)
      Substitute \( y = 4 \) into the line equation \(x - 2y + 4 = 0\) , we get
      \( x - 2(4) + 4 = 0\)
      \( x=4 \)
      So, the point of contact is \( (4,4) \)

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      (ii) perpendicular to the line \(2x-y=4\)
      Also find the point of contact.

      
      The given parabola is
      \(y^2=4x\) (i)
      Thus,
      \( 4a = 4\) gives \(a=1\)
      The equation of tangent to parabola, which is perpendicular to \(2x-y=4\) is
      \( x + 2y + k = 0 \) (A)
      \( 2y = -x - k \)
      \( y = -\frac{1}{2}x -\frac{k}{2} \) (ii)
      The condition for the line (ii) to be tangent to the parabola (i) is
      \(c=\frac{a}{m}\)
      \(-\frac{k}{2} =\frac{1}{-\frac{1}{2}}\)
      \(k=4\)
      Hence, the required equation of tangent line is, from (A)
      \( x + 2y + 4 = 0 \)
      Next, to find the point of contact, we substitute \(x = - 2y - 4\) into the parabola , so that
      \( y^2 = 4 \left(-2y - 4 \right) \)
      \( y^2 = -8y - 16 \)
      \( y^2 + 8y +16 = 0 \)
      \( y =- 4 \)
      Substitute \( y = -4 \) into the line equation \(x + 2y + 4 = 0\) , we get
      \( x + 2(-4) + 4 = 0\)
      \( x=4 \)
      So, the point of contact is \( (4,-4) \)
   2. Find the rquation of normal line to the parabola \(y^2=3x\)
      (i) parallel to the line \(y=2x+1\)
      (ii) perpendicular to the line \(3x+2y=8\)
      

      Answer 👉 Click Here

      (i) parallel to the line \(y=2x+1\)
      

      
      The given parabola is
      \(y^2=3x\) (i)
      Thus,
      \( 4a = 3\) gives \(a=\frac{3}{4}\)
      The equation of normal to parabola, which is parallel to \(y=2x+1\) is
      \( y=2x+k \) (A)
      The condition for the line (A) to be normal to the parabola (i) is
      \(c=-2am-am^3\)
      \(k=-2\frac{3}{4}. 2-\frac{3}{4} (8)\)
      \(k=-9\)
      Hence, the required equation of tangent line is, from (A)
      \( y=2x-9 \)
      

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      (ii) perpendicular to the line \(3x+2y=8\)
      

      
      The given parabola is
      \(y^2=3x\) (i)
      Thus,
      \( 4a = 3\) gives \(a=\frac{3}{4}\)
      The equation of normal to parabola, which is perpendicular to \(3x+2y=8\) is
      \( 2x-3y+k=0 \) (A) \( 3y=2x+k \)
      \( y=\frac{2}{3}x+\frac{k}{3} \) (ii)
      The condition for the line (ii) to be normal to the parabola (i) is
      \(c=-2am-am^3\)
      \(\frac{k}{3} =-2\frac{3}{4}. \frac{2}{3} -\frac{3}{4} (\frac{8}{27})\)
      \(k=-\frac{11}{3}\)
      Hence, the required equation of tangent line is, from (A)
      \( 3y=2x-\frac{11}{3} \)
      \( 9y=6x-11\)
      
4. Solve the followings
   1. Show that tangent to the parabola \(y^2=4x\) and \(x^2=4y\) at (1,2) and (-2,1) respectively are at right angles.
      
      
      Slope of the tangent to the Parabola \( y^2 = 4x \) at \( (1, 2) \)is
      \( m_1 = \frac{y_1}{2a} \)
      \( m_1 = \frac{2}{2} = 1 \)
      Next, slope of the tangent to the Parabola \( x^2 = 4y \) at \( (-2, 1) \)is
      \( m_2 = \frac{x_1}{2a} \)
      \( m_2 = \frac{-2}{2} = -1 \)
      The tangents are perpendicular if the product of their slopes is \(-1\).
      \( m_1 \times m_2 = 1 \times -1 = -1 \)
      Since the product of the slopes is \(-1\), the tangents are at right angles
   2. Show that normal to the parabola \(y^2=8x\) at (2,4) meets the parabola again in (18,-12)
      
      
      The equation of the parabola is \( y^2 = 8x \), so the slope of normal to the parabola is
      \( m= -\frac{2a}{y_1} \)
      \( m= - \frac{2.2}{4} = -1 \)
      Now, equation of the Normal at \( (2, 4) \) is
      \( y - y_1 = m(x - x_1) \)
      \( y - 4 = -1(x - 2) \)
      \( y = -x + 6 \)
      \( x + y - 6 = 0 \)
      To find where the normal meets the parabola again, substitute the equation of the normal \( x + y - 6 = 0 \) into the equation of the parabola \( y^2 = 8x \), we get
      \( y^2 = 8x \)
      \( (6 - x)^2 = 8x \)
      \( 36 - 12x + x^2 = 8x \)
      \( x^2 - 20x + 36 = 0 \)
      \( x = 2 \) or \( x = 18 \)
      When \( x = 2 y = 4 \) is the original point, and when \( x = 18 \), we get \( y = 6 - 18 = -12 \).
      Therefore, the normal meets the parabola again at \( (18, -12) \)
5. Solve the followings
   1. Find the equation of tangent to the parabola \(y^2=6x\) making angle 45 degree with the x-axis. Also find the point of contact.
      
      
      Given that, parabola is \(y^2=6x\), thus,
      \( 4a = 6\) gives \(a=\frac{3}{2}\)
      Also, given that tangent makes an angle of 45 degrees with the x-axis. Therefore, the slope \(m\) of the tangent is
      \( m = \tan(45^\circ) = 1 \)
      Now, the equation of the tangent to the parabola \( y^2 = 4ax \) is
      \( y=mx+\frac{a}{m}\)
      \( y = 1.x + \frac{3}{2} \)
      \( y = x + \frac{3}{2} \)
      To find the point of contact, we substitute the tangent equation \( y = x +\frac{3}{2}\) into the parabola equation \( y^2 = 6x \), so we get
      \( y^2 = 6x \)
      \( (x + \frac{3}{2})^2 = 6x \)
      \( (2x + 3)^2 = 24x \)
      \( (2x - 3)^2 = 0 \)
      \( x= \frac{3}{2}\) gives \( y = \frac{3}{2} + \frac{3}{2} =3\)
      Hence, the point of contact is \( (\frac{3}{2},3)\).
   2. A tangent to the parabola \(y^2=12x\) makes an angle 45 degree with the straight line \(2y=x+3\). Find its equation and point of contact.
   3. Find the equation of tangents to the parabola \(y^2=12x\) drawn through the points (-1,2). Also find the point of contact and the angle between two tangents.
   4. Find the equation of common tangents to the parabolas \( y^2=4x\) and \( x^2=4y\)
6. Solve the followings
   1. Show that the pair of tangents drawn from the point (-2,3) to the parabola \(y^2=8x\) are at right angle
   2. Prove that the tangents at the extrimities of the latus rectum of a parabola \(y^2=16x\) are at right angles

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