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
${\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{4}\right)}^2=2\Rightarrow a=4$