Q. The equation of the normal to the curve $y^{4} = ax^{3}\$at (a, a) is
A) $x + 2y = 3a$
B) $3x - 4y + a = 0$
C) $4x + 3y = 7a$
D) $4x - 3y = 0$
Explanation
(c)
On differentiating given curve w. r. t. $x$, we get
$4\ y^{3}\frac{dy}{dx} = 3ax^{2}$
$\Rightarrow \left( \frac{dy}{dx} \right)_{(a,\ a)} = \frac{3a^{3}}{4a^{3}} = \frac{3}{4}$
$\therefore$ Equation of normal at point $(a,\ a)$ is
$y - a = - \frac{4}{3}\ (x - a)$
$\Rightarrow \ \ 4x + 3y = 7a$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.