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

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

Hint:

Use the slope of the tangent.

Given:

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 $0$.

Now,

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