The normal at left( {2,frac{3}{2}} right | vedclass

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Question

Ellipsa is $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$

Now, equation of normal at $(2,3/2)$ is

$\frac{{16x}}{2} - \frac{{3y}}{{3/2}} = 1

Vedclass · Accepted Answer

$y^2 = -104 x$

Vedclass · Answer

Ellipsa is $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$

Now, equation of normal at $(2,3/2)$ is

$\frac{{16x}}{2} - \frac{{3y}}{{3/2}} = 16 - 3$

$ \Rightarrow 8x - 2y = 13$

$ \Rightarrow y = 4x - \frac{{13}}{2}$

Let $y = 4x - \frac{{13}}{2}$touches a parabola

      ${y^2} = 4ax$

We konw, a straight line $y=mx+c$ touches

a parabola ${y^2} = 4ax$ if $a-mc=0$

$	herefore a - \left( 4 ight)\left( { - \frac{{13}}{2}} ight) = 0 \Rightarrow a =  - 26$ 

Hence, required equation of parabola is

${y^2} = 4\left( { - 26} ight)x =  - 104x$

The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is

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