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
$\therefore \frac{-4x}{\alpha y}\times\frac{16}{3y^{2}}=-1 \Rightarrow 3\alpha y^{3}=64x$
$\Rightarrow \alpha=\frac{64x}{3\times16x}=\frac{4}{3}$