Normal at left( {2,frac{3}{2}} right) to ellipse, frac{{{x^2}}}{{16}} + frac{{{y^2

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10-2. Parabola, Ellipse, Hyperbola

hard

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

$y^2 = -104 x$

$y^2 = 14x$

$y^2 = 26x$

$y^2 = -14x$

(AIEEE-2012)

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 $a-mc=0$

$\therefore a – \left( 4 \right)\left( { – \frac{{13}}{2}} \right) = 0 \Rightarrow a = – 26$

Hence, required equation of parabola is

${y^2} = 4\left( { – 26} \right)x = – 104x$

Standard 11

Mathematics

normal

hard

(JEE MAIN-2019)

medium

hard

medium