Question

If the curves $\frac{x^2}{c}+\frac{y^2}{4}=1$ and $y^3=16 x$ intersect at right angles, then $c$ is equal to

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

Answer 1

Answer: (C)

Let $\left(x_1, y_1\right)$ be their point of intersection.
So, $m _1 \times m _2=-1 \Rightarrow\left(\frac{64 x _1}{ c }\right)=3 y _1{ }^3$...(1)
Also, $ y _1{ }^3=16 x _1$....(2)
$\therefore$ From (1) and (2), we get $c=\frac{4}{3}$