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  • Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 15 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter15 Tangents and Normals Exercise Fill in the blanks Question 15 Maths Textbook Solution.

Answers (1)

Answer:

               Point: \left ( 1,0 \right )

Hint:

  1. First we will find the slope of line x-y= 0 then evaluate with the slope of curve
  2. If tangent is perpendicular to the line then  m_{1} \cdot m_{2}=-1

Given:

Given curve  y=x^{2}-3 x+2  where the tangent is perpendicular to the line  x-y= 0

To find:

We have to find the co-ordinate of the point on the given curve.

Solution: 

We have,

               \begin{aligned} & x-y=0 \\ \Rightarrow \quad & y=x \end{aligned}                                                                                                                                                             … (i)

Comparing equation (i) with equation  y=m x+c

 \Rightarrow \quad m_{1}=1

Again, if the line is perpendicular to the tangent then

 \begin{aligned} &\Rightarrow \quad m_{1} \cdot m_{2}=-1 \\ &\Rightarrow \quad 1 \cdot m_{2}=-1 \end{aligned}                                                                                                                                          \left[\because m_{1}=1\right]

\Rightarrow \quad m_{2}=-1

Given curve,

         y=x^{2}-3 x+2                                                                                                                                                  … (ii)

On differentiating both side with respect to x, we get

\Rightarrow \quad \frac{d y}{d x}=2 x-3                                                                                                                        \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

Here slope of tangent is -1

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

Substituting the value of x in equation (ii), we get

          \begin{aligned} &y=1-3+2 \\ &y=0 \end{aligned}

Hence the point is \left ( x,y \right )= \left ( 1,0 \right )

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