Question

The equation of normal $x^{2/3}+y^{2/3}=a^{2/3}$ at $(\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}})$ is ______ .

• A. $2x+y=0$
• B. $y=1$
• C. $x=0$
• D. $x=y$

Answer 1

The correct option is D $x=y$

To get the slope of $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
We differentiate w.r.to $x$
$\therefore \frac{2}{3}{x}^{-\frac{1}{3}}+\frac{2}{3}{y}^{-\frac{1}{3}}\frac{dy}{dx}=0$
$\therefore \frac{dy}{dx}=-\left(\frac{x^{-\frac{1}{3}}}{y^{\frac{2}{3}}}\right)$
$\therefore \frac{dy}{dx}=-{\left(\frac{y}{x}\right)}^{\frac{1}{3}}$
$\therefore$ Slope of tangent at point
$P(\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}})$
$\therefore$ Slope $m={\left(\frac{dy}{dx}\right)}_{\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}}}$
$\therefore m={\left(\frac{dy}{dx}\right)}_{\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}}}-{\left(\frac{\frac{a}{2\sqrt{2}}}{\frac{a}{2\sqrt{2}}}\right)}^{\frac{1}{3}}$
$=-{\left(1\right)}^{\frac{1}{3}}=-1$
Hence slope of tangent $m=-1$
$\therefore$ Slope of normal $=1$
$\therefore$ we get equation of normal using equation $y-y_1=m(x-x_1)$
$\therefore y-\frac{a}{2\sqrt{2}}=\left(1\right)(x-\frac{a}{2\sqrt{2}})$
$\therefore y-\frac{a}{2\sqrt{2}}=x-\frac{a}{2\sqrt{2}}$
$\therefore$ $y=x$