Question

If the sum of squares of the intercept on the axes cut off by the tangent on the curve $x^{\frac{1}{3}}+y^{\frac{1}{3}}=a^{\frac{1}{3}},a>0$ at $(\frac{a}{8},\frac{a}{8})$ is $2,$ then value of $a$ is

Answer 1

$x^{\frac{1}{3}}+y^{\frac{1}{3}}=a^{\frac{1}{3}}$
Differentiating w.r.t. $x$
$\Rightarrow \frac{1}{3}{x}^{-\frac{2}{3}}+\frac{1}{3}{y}^{-\frac{2}{3}}\left(\frac{dy}{dx}\right)=0$
$\Rightarrow \frac{dy}{dx}=-\frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}$
At $(\frac{a}{8},\frac{a}{8}),\frac{dy}{dx}=-1$
Equation of tangent is
$y-\frac{a}{8}=-1(x-\frac{a}{8})$
$\Rightarrow x+y=\frac{a}{4}$
$x-$intercept $=\frac{a}{4}=y-$intercept
$\therefore \frac{a^2}{16}+\frac{a^2}{16}=2$
$\therefore a^2=16$
$\therefore a=4$