Question

If the curves $\frac{x^2}{\alpha }+\frac{y^2}{4}=1$ and $y^3=16x$ intersect at right angles, then a value of $\alpha$ is :

• A. $2$
• B. $\frac{4}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

$\frac{x^2}{\alpha }+\frac{y^2}{4}=1\Rightarrow \frac{2x}{\alpha }+\frac{2y}{4}.\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-4x}{\alpha y}...\left(i\right)$
$y^3=16x\Rightarrow 3y^2.\frac{dy}{dx}=16\Rightarrow \frac{dy}{dx}=\frac{16}{3y^2}...\left(ii\right)$
Since curves intersects at right angles
$\therefore \frac{-4x}{\alpha y}\times \frac{16}{3y^2}=-1\Rightarrow 3\alpha y^3=64x$
$\Rightarrow \alpha =\frac{64x}{3\times 16x}=\frac{4}{3}$