Question

The equation normal to the curve $x^{2/3}+y^{2/3}=a^{2/3}$ at the point $(a,0)$ is-

• A. $x=a$
• B. $x=-a$
• C. $y=a$
• D. $y=-a$

Answer 1

Answer: (A)

$x=a$

We have,

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

Then,

$\frac{d}{dx}(x^{\frac{2}{3}}+y^{\frac{2}{3}})=\frac{d}{dx}\left(a^{\frac{2}{3}}\right)$

$\frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0$

$x^{-\frac{1}{3}}+y^{-\frac{1}{3}}\frac{dy}{dx}=0$

$\frac{dy}{dx}={(-\frac{x}{y})}^{-\frac{1}{3}}$

$\frac{dy}{dx}={(-\frac{y}{x})}^{\frac{1}{3}}$

At point $(a,0)$

${\left(\frac{dy}{dx}\right)}_{(a,0)}={{(-\frac{y}{x})}^{\frac{1}{3}}}_{(a,0)}$

${\left(\frac{dy}{dx}\right)}_{(a,0)}={(-\frac{0}{a})}^{\frac{1}{3}}$

${\left(\frac{dy}{dx}\right)}_{(a,0)}=0$

Equation of normal is

$y-y_1=-\frac{1}{{\left(\frac{dy}{dx}\right)}_{(a,0)}}(x-x_1)$

$y-y_1=\frac{1}{0}(x-x_1)$

Equation of normal at point $(a,0)$ and we get,

$y-0=\frac{1}{0}(x-a)$

$x-a=0$

$x=a$

Hence, this is the answer.