Question

# If the sum of the squares of the intercepts on the axes cut off by the tangent to the cuve $$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$$ at $$\displaystyle \left ( \dfrac{a}{8}, \dfrac{a}{8} \right )$$ is $$2$$, then $$a$$ has the value

A
1
B
2
C
4
D
8

Solution

## The correct option is C $$4$$$$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$$$$y'(x)= -\cfrac{y^{2/3}}{x^{2/3}}$$$$\left.y'\right|_{(x,y)=\left(\dfrac{a}{8}, \dfrac{a}{8}\right)} = -1$$Equation of tangent$$x+y =c$$$$x +y = \dfrac{a}{4};\>\>\>\>\>\> \because \left(\dfrac{a}{8}, \dfrac{a}{8}\right)$$ satisfies it$$x-$$intercept $$= a\dfrac{a}{4} = y-$$interceptGiven, $$\cfrac{a^2}{16}+\cfrac{a^2}{16} = 2$$$$a = \pm 4$$$$a =4;\>\>\>\>\>\because a>0$$Mathematics

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