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. $y^2=14\,x$
• C. $y^2 = 26\,x$
• D. $y^2 = - 14\,x$

Answer 1

Answer: (A)

$y^2 =- 10^4\,x$

Answer 2

Ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{3} = 1$ Now, equation of normal at $\left(2, 3/2\right)$ is $\frac{16x}{2}-\frac{3y}{3/2} = 16-3$ $\Rightarrow\quad8x-2y= 13$ $\Rightarrow \quad y=4x-\frac{13}{2}$ 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 $= 0$ $\therefore\quad a-\left(4\right)\left(-\frac{13}{2}\right) = 0 \Rightarrow a = -26$ Hence, required equation of parabola is $y^{2} = 4\left(-26\right)x = -104\,x$