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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
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B
2
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C
4
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D
8
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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 $$= a\dfrac{a}{4} = y-$$intercept
Given, $$ \cfrac{a^2}{16}+\cfrac{a^2}{16} = 2 $$
$$a =  \pm 4$$
$$a =4;\>\>\>\>\>\because a>0$$

Mathematics

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