Question

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 constant.

Answer 1

Equation of the curve is $x^{2/3}+y^{2/3}=a^{2/3}$

the parametric equations of the curve are $x=a{\cos }^3\theta$,

$y=a{\sin }^3\theta$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\cdot 3{\sin }^2\theta \cos \theta }{a\cdot 3{\cos }^2\theta (-\sin \theta )}$

$=-\tan \theta$

Equation of the tangent at $(a{\cos }^3\theta ,a{\sin }^3\theta )$ is $y-a{\sin }^3\theta =-\tan \theta (x-a{\cos }^2\theta )$

$y-a{\sin }^3\theta =-x\tan \theta +a\frac{\sin \theta }{\cos \theta }{\cos }^3\theta$

$y-a{\sin }^3\theta =-x\frac{\sin \theta }{\cos \theta }+a\sin \theta {\cos }^2\theta$

$\Rightarrow y\cos \theta +x\sin \theta =a\sin \theta {\cos }^3\theta +a\frac{{\sin }^3\theta }{\cos \theta }$

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

$\Rightarrow \frac{x}{\cos \theta }+\frac{y}{\sin \theta }=a$ ( Divided by both side $\sin \theta \cos \theta )$

x-intercept $=a\cos \theta$, y-intercept$=a\sin \theta$

$A(a\cos \theta ,0)$ and $B(0,a\sin \theta )$

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

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

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

$\therefore AB=a$, a constant.