Question:
The point on the curve $y^{2}=x$ where tangent makes $45^{\circ}$ angle with $x$-axis is
A. $\left(\frac{1}{2}, \frac{1}{4}\right)$
B. $\left(\frac{1}{4}, \frac{1}{2}\right)$
C. $(4,2)$
D. $(1,1)$
Solution:
Given that $\mathrm{y}^{2}=\mathrm{x}$
The tangent makes $45^{\circ}$ angle with $\mathrm{x}$-axis.
So, slope of tangent $=\tan 45^{\circ}=1$
$\because$ the point lies on the curve
$\therefore$ Slope of the curve at that point must be 1
$2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}$
$\Rightarrow \frac{1}{2 \mathrm{y}}=1$
$\Rightarrow \mathrm{y}=\frac{1}{2}$
And $\mathrm{x}=\frac{1}{4}$
So, the correct option is B.