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  • Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 8 maths text book solution.

Provide solution for RD Sharma maths class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 8 maths text book solution.

Answers (1)

Answer : (x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)

Hint : \text { slope of tangent }=\tan \frac{\pi}{4}=1

Given :

Here the given curve, y^{2}=x where the tangent line makes an angle \frac{\pi }{4} with x- axis.

We have to find the co-ordinate of the point on the given curve.

Solution:

Let the required point be (x,y).

We know,

The tangent makes an angle 45^{o}with the x-axis.

\therefore              \text { slope of tangent }=\tan 45^{\circ}=1

Since the point lies on the curve

Hence,  y_{1}^{2}=x_{1}

Now,

          y^{2}=x

Differentiating with respect to x,

\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=1 \\ \Rightarrow & \frac{d y}{d x}=\frac{1}{2 y} \end{array}

\therefore \quad \text { slope of tangent }=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1}{2 y_{1}}

\Rightarrow \quad \frac{1}{2 y_{1}}=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \text { slope of tangent at angle } \frac{\pi}{4}=\tan \frac{\pi}{4}=1\right]

\Rightarrow \quad 2 y_{1}=1

\Rightarrow \quad y_{1}=\frac{1}{2}

Now, we have,

\begin{aligned} &x_{1}=y_{1}^{2} \\ &\Rightarrow x_{1}=\left(\frac{1}{2}\right)^{2} \\ &\Rightarrow \quad x_{1}=\frac{1}{4} \end{aligned}

Hence, (x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)

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