Question

If curves $\frac{x^2}{a^2}+\frac{y^2}{4}=1$ and $y^3=16x$ intersect orthogonally, then $a^2$ equals:

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

Answer 1

Answer: (B)

Given, curves $\frac{x^2}{a^2}+\frac{y^2}{4}=\mathrm{1.....}\left(i\right)$ and $y^3=16x.....\left(ii\right)$ From $\left(i\right),\frac{dy}{dx}=-\frac{x}{a^2}\frac{4}{y}$ , From $\left(ii\right),\frac{dy}{dx}=-\frac{16}{3y^2}$
$\because$ Curves intersect orthogonally
$\therefore$ $\left(-\frac{4}{a^2}\frac{x}{y}\right)\left(\frac{16}{3y^2}\right)=-1$
$\Rightarrow -\frac{64x}{3a^2{y}^3}=-1$
$\Rightarrow \frac{4}{3a^2}=1$ $\left[\because y^3=16x\right]$
$\Rightarrow a^2=\frac{4}{3}$