If the tangent at any point on the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ intersects the coordinate axes in $A$ and $B$, then show that the length $AB$ is a constant.

Given curve is $x^{2/3}+y^{2/3}=a^{2/3}\to \left(1\right)$

Let $P\left(a{\cos }^3\theta ,a{\sin }^3\theta \right)$ is a point on the curve

Let $x=a{\cos }^3\theta ;y=a{\sin }^3\theta$

Diff. w.r.t. $\theta$

$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta ;\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$

Slope $m=\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }$

$m=\frac{-\sin \theta }{\cos \theta }$

Eq. of tangent at $P$ is $y-y_1=m\left(x-x_1\right)$

$\Rightarrow y-a{\sin }^3\theta =\frac{-\sin \theta }{\cos \theta }\left(x-a{\cos }^3\theta \right)$

$\Rightarrow \frac{1}{\sin \theta }\left(y-a{\sin }^3\theta \right)=\frac{-x}{\cos \theta }+a{\cos }^2\theta$

$\Rightarrow \frac{y}{\sin \theta }-a{\sin }^2\theta =\frac{-x}{\cos \theta }+a{\cos }^2\theta$

$\Rightarrow \frac{x}{\cos \theta }+\frac{y}{\sin \theta }=a{\cos }^2\theta +a{\sin }^2\theta$

$\Rightarrow \frac{x}{\cos \theta }+\frac{y}{\sin \theta }=a\Rightarrow \frac{x}{a\cos \theta }+\frac{y}{a\sin \theta }=1$

This tangent intersect $X$-axis at $A(a\cos \theta ,0)$ and $Y$-axis at $B(0,a\sin \theta )$

Now $AB=\sqrt{(a\cos \theta -0{)}^2+(0-a\sin \theta {)}^2}$

$=\sqrt{a^2\left({\cos }^2+{\sin }^2\theta \right)}$

$=\sqrt{a^2}=a$