Q. The point on the curve $y^{2} = x$, the tangent at which makes an angle $45{^\circ}$ with $x$-axis is
A) $\left( \frac{1}{4},\frac{1}{2} \right)$
B) $\left( \frac{1}{2},\frac{1}{4} \right)$
C) $\left( \frac{1}{2},\ - \frac{1}{2} \right)$
D) $\left( \frac{1}{2},\frac{1}{2} \right)$
Explanation
(a)
Given, $y^{2} = x$ …(i)
$\Rightarrow \ \ \ \ \ \ \ \ 2y\frac{du}{dx} = 1$
$\Rightarrow \ \ \ \ \ \ \frac{dy}{dx} = \frac{1}{2y} =$slope
Also, slope $\tan{45{^\circ}} = 1$
$\Rightarrow \ \ \ \ \ \ \frac{1}{2y} = 1$
$\Rightarrow \ \ \ \ \ \ y = \frac{1}{2}$
From Eq. (i), if $y = \frac{1}{2}$, then $x = \frac{1}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.