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The equation to the normal to the curve $y^{4} = a x^{3}$ at (a,a) is

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4x-3y=0

3x-4y+a=0

x+2y=3a

4x+3y=7a

answer is C.

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Detailed Solution

$y^{4} = a x^{3} \Rightarrow 4 y^{3} \frac{d y}{d x} = 3 a x^{2}$

$\Rightarrow \frac{d y}{d x} = \frac{3 a x^{2}}{4 y^{3}}$

$m = (\frac{d y}{d x}) (a, a) = \frac{3 a^{3}}{4 a^{3}} = \frac{3}{4}$

Equation of the normal is

$y - a = \frac{- 4}{3} (x - a) \Rightarrow 4 x + 3 y = 7 a$

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