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The point on the curve


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. $(1,1)$


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.

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