Question

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

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

Answer 1

Answer: (A)

$x=a$

Given curve is,
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
Differentiating w.r.t $x$
$\frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}={\left(\frac{y}{x}\right)}^{\frac{1}{3}}$
Thus slope of normal at $(a,0)$ is $m=-\frac{dx}{dy}={\left(\frac{a}{0}\right)}^{\frac{1}{3}}\to \infty$
Therefore, equation of normal is, $x=a$
Hence, option 'A' is correct.