The point on the curve

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 $y^{2}=x$

The tangent makes $45^{\circ}$ angle with $x$-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

$2 y \frac{d y}{d x}=1$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$

$\Rightarrow \frac{1}{2 y}=1$

$\Rightarrow y=\frac{1}{2}$

And $x=\frac{1}{4}$

So, the correct option is B

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