Question

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

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (C)

$x^{1/3} + y^{1/3} = a^{1/3} \Rightarrow a^{1/3}$
$ \Rightarrow \frac{dy}{dx} = - \frac{x^{-2/3}}{y^{-2/3}} = - \frac{x^{2/2}}{y^{2/3}} $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(a/8, a/8\right)} = - 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 $