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Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 1 maths textbook solution.

Answers (1)

Answer :

\left(x_{1}, y_{1}\right)=(1,2)


Use the slope of the tangent.


Here the curve y=x^{2}-2 x+3, where the tangent is parallel to x -axis.

We have to find the point on the given curve.

Solution :

The slope at the x -axis is 0.


Let  \left ( x_{1},y_{1} \right ) be the required point.

Since the point lies on the curve.

Hence y_{1}=x_{1}^{2}-2 x_{1}+3                                                       ....(i)

Then, y=x^{2}-2 x+3

\Rightarrow \quad \frac{d y}{d x}=2 x-2

Slope of the tangent at (x, y)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}

                                                  = 2x_{1}-2

Here \frac{d y}{d x}=0

\begin{aligned} &\Rightarrow \quad 2 x_{1}-2=0 \\ &\Rightarrow \quad x_{1}=1 \end{aligned}

From equation (i)

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

Hence required point is \left ( x_{1},y_{1} \right )= (1,2)

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