#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 16 Maths Textbook Solution.

Answer: The requried points are (3,2)

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

Given:

$\text { Equation of the curve is } y=x^{2}-4 x+5 \quad \rightarrow(1)$

$\text { Equation of the line is } 2 y+x=7 \quad \rightarrow(2)$

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

$\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0$

$\frac{d y}{d x}=2 x^{2-1}-4$

$\frac{d y}{d x}=2 x-4$

$\text { Slope } \mathrm{m}_{1}=\frac{d y}{d x}=2 x-4 \quad \rightarrow(3)$

$\text { Differentiating eqn }(2) \text { with respect to } x$

$2 \frac{d y}{d x}+1=0$

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

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

$\text { Slope } \mathrm{m}_{2}=\frac{d y}{d x}=\frac{-1}{2}$

$\text { We have given that slope(1) and slope (2) are perpendicular to each other }$

Hence,

\begin{aligned} &m_{1} \times m_{2}=-1 \\ &(2 x-4) \times\left(\frac{-1}{2}\right)=-1 \\ &(2 x)\left(\frac{-1}{2}\right)-(4)\left(\frac{-1}{2}\right)=-1 \end{aligned}

\begin{aligned} &-x+2=-1 \\ &x=2+1 \\ &x=3 \end{aligned}

$\text { When put } x=3 \text { in eqn (1) } y=x^{2}-4 x+5$

\begin{aligned} &y=(3)^{2}-4(3)+5 \\ &y=9-12+5 \\ &y=2 \end{aligned}

Thus the required points are (3,2)