Question

If $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$, then $\frac{d y}{d x}$ is equal to

• A. $-\sqrt[3]{\frac{y}{x}}$
• B. $\sqrt[3]{y / x}$
• C. $\frac{y}{x}$
• D. None of these

Answer 1

Answer: (A)

Let $x=a \cos ^3 \theta, y=a \sin ^3 \theta$. Then,
$ x^{\frac{2}{3}}+y^{\frac{2}{3}}=\left(a \cos ^3 \theta\right)^{\frac{2}{3}}+\left(\operatorname{asin}^3 \theta\right)^{\frac{2}{3}}$
$ =a^{\frac{2}{3}}\left[\left(\cos ^2 \theta+\left(\sin ^2 \theta\right)\right]=a^{\frac{2}{3}}\right.$
Hence, $x=\operatorname{acos}^3 \theta, y=\operatorname{cosin}^3 \theta$ is parametric equation of $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Now, $\frac{d x}{d \theta}=-3 a \cos ^2 \theta \sin \theta$ and $\frac{d y}{d \theta}=3 a \sin ^2 \theta \cos \theta$
$\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}=-\tan \theta=-\sqrt[3]{\frac{y}{x}}$