Question

If the tangent at any point on the curve $x^4+y^4=a^4$ cuts off intercepts p and q on the coordinate axes, the value of $p^{-\frac{4}{3}}+q^{-\frac{4}{3}}$ is

• A. $a^{-\frac{4}{3}}$
• B. $a^{-\frac{1}{2}}$
• C. $a^{\frac{1}{2}}$
• D. none of these

Answer 1

The correct option is A $a^{-\frac{4}{3}}$

Let
$x=a\sqrt{\cos \theta }$
$y=a\sqrt{\sin \theta }$
Hence
$\frac{dy}{dx}$

$=-\frac{\cos \theta .\sqrt{\cos \theta }}{\sin \theta .\sqrt{\sin \theta }}$
$=\cot {\theta }^{\frac{3}{2}}$.
Hence equation of tangent will be
$y-a\sqrt{\sin \theta }=-\frac{\cos \theta .\sqrt{\cos \theta }}{\sin \theta .\sqrt{\sin \theta }}(x-a\sqrt{\cos \theta })$
$\sin \theta .\sqrt{\sin \theta }y-a{\sin }^2\theta =-\cos \theta \sqrt{\cos \theta }x+a{\cos }^2\theta$
$(\sin \theta {)}^{\frac{3}{2}}.y+(\cos \theta {)}^{\frac{3}{2}}x=a$
Hence the intercepts are
$Q=\frac{a}{(\sin \theta {)}^{\frac{3}{2}}}$ and $P=\frac{a}{(\cos \theta {)}^{\frac{3}{2}}}$
Hence
$Q^{\frac{-4}{3}}+P^{\frac{-4}{3}}$
$=a^{-\frac{4}{3}}[{\cos }^2\theta +{\sin }^2\theta ]$
$=a^{-\frac{4}{3}}$.