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