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Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 18

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        \left ( \frac{1}{4} ,\frac{1}{2}\right )


First find the slope of curve  y^{2}=x, then compare with slope 45^{\circ} with x-axis.

           i.e. slope  \tan\frac{\pi}{4}=1


Given curve,  y^{2}=x, the tangent at which makes an angle of 45^{\circ} with x-axis.

To find:

We have to find the point on the given curve.


We have

        y^{2}=x                                                                                                                                                                       … (i)

Differentiate both side with respect to x, we get

\Rightarrow   2y\frac{dy}{dx}=1                                                                                                                              \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

\Rightarrow   \frac{dy}{dx}=\frac{1}{2y}                                                                                                                                                                … (ii)

Also tangent makes an angle of   45^{\circ}\left ( \frac{\pi}{4} \right )  with x-axis

\Rightarrow   \frac{dy}{dx}=\tan\left ( \frac{\pi}{4} \right )                                                                                                                             \left[\because \tan \left(\frac{\pi}{4}\right)=1\right]

\Rightarrow   \frac{dy}{dx}=1                                                                                                                                                                  … (iii)

From equation (ii) and (iii), we get

\begin{aligned} &\Rightarrow \quad \frac{1}{2 y}=1 \\ &\Rightarrow \quad y=\frac{1}{2} \end{aligned}

Put value of  y in equation  y^{2}=x

\Rightarrow         x=\frac{1}{4}

Hence required point is \left ( \frac{1}{4} ,\frac{1}{2}\right )

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