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

Answers (1)


                a+b= 10


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


The given equation of curve,

             y^{2}=5 x-1  

To find:

We have to find the value of  a+b


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


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