Q106 / 160 · AP EAMCET 2008 · Mathematics

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

  • A
    $x+2 y=3 a$
  • B
    $3 x-4 y+a=0$
  • Correct answer
    C
    $4 x+3 y=7 a$
  • D
    $4 x-3 y=0$

Solution

Given curve is $y^4=a x^3$ On differentiating w.r.t. $x$, we get $$ \begin{aligned} & 4 y^3 \frac{d y}{d x}=3 a x^2 \\ & \Rightarrow \quad\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{3 a^3}{4 a^3}=\frac{3}{4} \\ & \end{aligned} $$ $\therefore$ Equation of normal at point $(a, a)$ is $$ \begin{array}{cc} & y-a=-\frac{4}{3}(x-a) \\ \Rightarrow & 4 x+3 y=7 a \end{array} $$

Hence, the correct answer is option: 'C'.

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