Question

If a normal to ellipse $\frac{x^2}{18}+\frac{y^2}{8}=1$ at $(-3,-2)$ touches the parabola $y^2=4ax$, then value of a is

• A. $\frac{15}{4}$
• B. $15$
• C. $\frac{15}{2}$
• D. $\frac{-15}{4}$

Answer 1

Answer: (A)

$\frac{15}{4}$

The slope of the tangent to $\frac{x^2}{18}+\frac{y^2}{8}=1$ is found by

$\frac{2x}{18}+\frac{2y\frac{dy}{dx}}{8}=0$

$\frac{dy}{dx}=\frac{-4x}{9y}=\frac{-4\times -3}{9\times -2}=\frac{-2}{3}$

Thus, the slope of the normal would be $\frac{3}{2}$

The equation of normal thus becomes $y=\frac{3x}{2}+\frac{5}{2}$

(using slope and point formula for equation of straight line)

Substituting $x=\frac{2y-5}{3}$ in the parabola equation, we get

$y^2=4a\left(\frac{2y-5}{3}\right)$

$\Rightarrow 3y^2-8ay+20a=0$

This must have the determinant to be $0.$

$\Rightarrow 64a^2=4\left(3\right)\left(20a\right)$

Or,

$a=\frac{15}{4}$