#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 4 Maths Textbook Solution.

Answer:$x=\pm \sqrt{\frac{7}{3}} \& y=\pm \frac{-2}{3} \sqrt{\frac{7}{3}}$

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

Given:$\text { Curve } y=x^{3}-3 x$

Solution:

$\text { First we will find the slope of tangent, we get }$

$y=x^{3}-3 x$

$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}\right)-\frac{d}{d x}(3 x)$

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

$\frac{d y}{d x}=3 x^{2}-3 \quad \rightarrow(1)$

$\inline \text { The equation of line passing through }\left(x_{0}, y_{0}\right) \text { and the slope } m \text { is } y-y_{0}=m\left(x-x_{0}\right)$

$\inline \text { So the slope } m=\frac{y-y_{0}}{x-x_{0}}$

$\inline \text { The slope of the chord joining }(1,-2) \&(2,2)$

$\inline \frac{d y}{d x}=\frac{2-(-2)}{2-1}=\frac{4}{1}$

$\inline \frac{d y}{d x}=4 \quad \rightarrow(2)$

$\inline \text { from (1) \&(2) }$

$\inline 3 x^{2}-3=4$

$\inline 3 x^{2}=4+3$

$\inline 3 x^{2}=7$

$\inline x^{2}=\frac{7}{3}$

$\inline x=\pm \sqrt{\frac{7}{3}}$

$\inline y=x^{3}-3 x$

$\inline y=x\left(x^{2}-3\right)$

$\inline y=\pm \sqrt{\frac{7}{3}}\left(\left(\pm \sqrt{\frac{7}{3}}\right)^{2}-3\right)$

$\inline y=\pm \sqrt{\frac{7}{3}}\left(\frac{7}{3}-3\right)$

$\inline y=\pm \sqrt{\frac{7}{3}}\left(\frac{7-9}{3}\right)$

$\inline y=\pm \sqrt{\frac{7}{3}}\left(\frac{-2}{3}\right)$

$\inline y=\pm\left(\frac{-2}{3}\right) \sqrt{\frac{7}{3}}$

$\inline \text { Thus, the required point is } x=\pm \sqrt{\frac{7}{3}} \& y=\pm \frac{-2}{3} \sqrt{\frac{7}{3}}$