Question

Find the equations of the tangent and normal to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2\text{at}(1,1)$

Answer 1

Given curve,
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$

Differentiating w.r.t. $x,$

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

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

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

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

Slope of tangent at $(1,1)$ is :

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

We know,

$\text{Slope of the normal}=-\frac{1}{\text{Slope of the tangent}}=1$

Now, equation of tangent at $(1,1)$ is $y-1=\frac{dy}{dx}(x-1)$

$\Rightarrow y-1=-1(x-1)$

$\therefore x+y=2$

Equation of normal at $(1,1)$ is :

$\Rightarrow y-1=1(x-1)$

$\therefore y=x$

Hence, equation of tangent is

$y+x-2=0$

and equation of normal is

$y-x=0$.