If the curves x^2+y^24=1 and y^3=16x intersect at right angles, then a value of is:

Question

If the curves $\dfrac{x^2}{\alpha}+\dfrac{y^2}{4}=1$ and $y^3=16x$ intersect at right angles, then a value of $\alpha$ is:

Accepted Answer

For the ellipse: $\dfrac{dy}{dx}=-\dfrac{4x}{\alpha y}$. For $y^3=16x$: $\dfrac{dy}{dx}=\dfrac{16}{3y^2}$.

Right angles: $-\dfrac{4x}{\alpha y}\cdot\dfrac{16}{3y^2}=-1\Rightarrow3\alpha y^3=64x\Rightarrow\alpha=\dfrac{64x}{3\cdot16x}=\dfrac43$.

Solutions · 1