If the sum of the squares of the intercepts on the axes cut off by the tangent to the curve $x^{1 / 3}+y^{1 / 3}=a^{1 / 3}$ (with a > 0) at $(a / 8, a / 8)$ is 2, then a has the value
Answer & explanation
Correct answer: option 3
We have,
$x^{1 / 3}+y^{1 / 3}=a^{1 / 3}$
$\Rightarrow \frac{1}{3} x^{2 / 3}+\frac{1}{3} y^{-2 / 3} \frac{d y}{d x}=0 $
$\Rightarrow \frac{d y}{d x}=-\frac{x^{-2 / 3}}{y^{-2 / 3}}=-\frac{x^{2 / 3}}{y^{2 / 3}} \Rightarrow\left(\frac{d y}{d x}\right)_{(a / 8, a / 8)}=-1$
The equation of the tangent at $(a / 8, a / 8)$ is given by
$y-\frac{a}{8}=-1\left(x-\frac{a}{8}\right) \Rightarrow x+y-\frac{a}{4}=0$
The x and y intercepts of this line on the coordinate axes are each equal to a/4. So, we have
$\left(\frac{a}{4}\right)^2+\left(\frac{a}{4}\right)^2=2 \Rightarrow a=4$