#### Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 11

$a+b= 10$

Hint:

First find the equation of normal then comparing with $a x-b y+b=0$

Given:

The given equation of curve,

$y^{2}=5 x-1$

To find:

We have to find the value of  $a+b$

Solution:

We have,

$y^{2}=5 x-1$                                                                                                                              … (i)

Differentiating equation (i) with respect to $x$ , we get

$\Rightarrow \quad 2 y \frac{d y}{d x}=5$                                                                                                                        $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\frac{5}{2 y} \\\\ &\therefore \quad\left(\frac{d y}{d x}\right)_{(1,-2)}=\frac{-5}{4} \end{aligned}

Then the equation of normal at point $\left ( 1,-2 \right )$  is

$\begin{array}{ll} & (y-(-2))=\frac{-1}{\frac{-5}{4}}(x-1) \\\\ \Rightarrow \quad & 5(y+2)=4(x-1) \\\\ \Rightarrow \quad & 4 x-5 y-14=0 \end{array}$                                                                                                                       … (ii)

As the normal is of the form  $a x-5 y+b=0$                                                                                      … (iii)

Comparing equation (ii) and (iii), we get

$a= 4$   and $b= -14$

Hence   $a+b=-10$