Q. If $x = t^{2}$ and $y = 2t$, then equation of the normal at $t = 1$, is
A) $x + y - 3 = 0$
B) $x + y - 1 = 0$
C) $x + y + 1 = 0$
D) $x + y + 3 = 0$
Explanation
(a)
$\because\ \ \ x = t^{2}$ and $y = 2t$
$\therefore$ At $t = 1,\ \ x = 1$ and $y = 2$
Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{2t} = \frac{1}{t}$
$\Rightarrow \ \ \ \ \left( \frac{dy}{dx} \right)_{(1,\ \ 2)} = 2$
$\therefore$ Equation of normal is
$y - 2 = - 1(x - 1)$
$\Rightarrow \ \ \ \ \ \ x + y - 3 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.