Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 10 Maths Textbook Solution.

Answer:$\text { The required point is }\left(\frac{1}{2}, \frac{1}{4}\right)$

Hint: $\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

Given: $\text { The curve is } y=x^{2}$

Solution: $y=x^{2}$

$\text { Differentiating the above with respect to } x$

$\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}$

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

$\frac{d y}{d x}=2 x \quad \rightarrow(1)$

$\frac{d y}{d x}=\text { The slope of tangent }=\tan \theta$

$\text { Since the tangent make an angle of } 45^{\circ} \text { with } \mathrm{x} \text { -axis }$

$\frac{d y}{d x}=\tan \left(45^{\circ}\right)=1 \quad \rightarrow(2)$

From (1)&(2) we get

$2 x=1$

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

$\text { Substituting } x=\frac{1}{2} \text { in } y=x^{2} \text { , we get }$

$y=\left(\frac{1}{2}\right)^{2}$

$y=\frac{1}{4}$

So, the required point is $\left(\frac{1}{2}, \frac{1}{4}\right)$