Question

If the normal to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ makes an angle $\theta$ with $x$-axis then the slope of the normal is:

Answer 1

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

Differentiating with respect to $x$ give us

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

$x^{-\frac{1}{3}}+y^{\prime }.y^{\frac{-1}{3}}=0$

$x^{-\frac{1}{3}}=-y^{\prime }.y^{\frac{-1}{3}}$

$y^{\prime }={\left(\frac{y}{x}\right)}^{\frac{1}{3}}$

Therefore slope of normal

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

Since it makes an angle $\theta$ with the positive x axis,

$\tan \theta ={\left(\frac{-x}{y}\right)}^{\frac{1}{3}}$