Question

The point on the curve $y^2=x,$ the tangent at which makes an angle of ${45}^0$ with positive direction of $x-$ axis will be given by

• A. $\left(\frac{1}{2},\frac{1}{4}\right)$
• B. $\left(\frac{1}{2},\frac{1}{2}\right)$
• C. $(2,4)$
• D. $\left(\frac{1}{4},\frac{1}{2}\right)$

Answer 1

The correct option is D $\left(\frac{1}{4},\frac{1}{2}\right)$

Given equation of curve is $y^2=x$
$\frac{dy}{dx}=\frac{1}{2y}$
Slope of tangent $=\frac{1}{2y}$
Also, given slope of tangent $=tan{45}^{\circ }$
$\frac{1}{2y}=1\Rightarrow y=\frac{1}{2}$
$\Rightarrow x=\frac{1}{4}$
Hence, the required point is $(\frac{1}{4},\frac{1}{2})$