#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 21 Maths Textbook Solution.

Answer: The requried point is $(0,0) \text { or }\left(\frac{1}{3}, \frac{1}{27}\right)$

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

Given: $\text { The equation of the curve is } y=x^{3} \rightarrow(1)$

Solution:$\text { Differentiating eqn(1) with respect to } x$

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &\frac{d y}{d x}=3 x^{2} \rightarrow(2) \end{aligned}

$\text { Slope of the tangent to (1) is }$

$m_{1}=\frac{d y}{d x}=3 x^{2}$

$\text { Also, given that slope of the tangent is parallel to } \mathrm{x} \text { -coordinate of the point }$

$m_{2}=\frac{d y}{d x}=x \quad \rightarrow(3)$

From (2) & (3)

$m_{1}=m_{2}$

$3 x^{2}=x$

$3 x^{2}-x=0$

$x(3 x-1)=0$

$x=0 \text { or } 3 x-1=0$

$x=0 \text { or } x=\frac{1}{3}$

$\text { Substitute } x=0 \text { in } y=x^{3}$

$y=0$

$\text { Substitute } x=\frac{1}{3} \text { in } y=x^{3}$

$y=\left(\frac{1}{3}\right)^{3}$

$y=\frac{1}{27}$

$y=0 \text { or } \frac{1}{27}$

$\text { Thus the required point is }(0,0) \text { or }\left(\frac{1}{3}, \frac{1}{27}\right)$