Q58 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
The curves $\frac{x^2}{a^2}+\frac{y^2}{4}=4$ and $y^3=16 x$ intersect each other orthogonally, then $\mathrm{a}^2=$
- A2
- B$\frac{3}{4}$
- C$\frac{1}{2}$
- Correct answerD$\frac{4}{3}$
Solution
$$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\
& \therefore \frac{1}{a^2} 2 x+\frac{1}{4} 2 y \frac{d y}{d x}=0 \quad \Rightarrow \frac{d y}{d x}=\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)
\end{aligned}
$$
Also $\mathrm{y}^3=16 \mathrm{x}$
$$
\therefore 3 y^2 \frac{d y}{d x}=16 \Rightarrow \frac{d y}{d x}=\frac{16}{3 y^2}
$$
Since curves intersect orthogonally, from (1) and (2), we write
$$
\left(\frac{-x}{a^2}\right)\left(\frac{4}{y}\right)\left(\frac{16}{3 y^2}\right)=-1
$$
$\therefore \frac{64 x}{3 a^2 y^3}=1$ and we have $y^3=16 x$
$$
\therefore \frac{64 \mathrm{x}}{3 \mathrm{a}^2(16 \mathrm{x})}=1 \Rightarrow \mathrm{a}^2=\frac{4}{3}
$$
Hence, the correct answer is option: 'D'.