Question

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

• A. $y^2=-10^{4}x$
• B. $y^2=14x$
• C. $y^2=26x$
• D. $y^2=-14x$

Answer 1

Answer: (A)

Ellipse is $\frac{x^2}{16}+\frac{y^2}{3}=1$
Now, equation of normal at $\left(2,3/2\right)$ 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 know, a straight liney $=mx+c$ touches a parabola $y^2=4ax$ if a-me $=0$
$\therefore a-\left(4\right)\left(-\frac{13}{2}\right)=0\Rightarrow a=-26$
Hence, required equation of parabola is
$y^2=4\left(-26\right)x=-104x$