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  • Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 11 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 11 textbook solution.

Answers (1)

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

Hint :

For line, the slope of a line is denoted by m.

               y=m x+c, where m is slope of line.

Given :

Given that the curve, y^{2}=3-4 x

Where the tangent parallel to the line, 2 x+y-2=0

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

Solution :

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

We know that,

If  2 x+y-2=0

\Rightarrow \quad y=-2 x+2

Comparing with equation:

y=mx+c

We get, m=-2

Slope of the given line =-2

Since, the point lies on the curve

So         y_{1}^{2}=3-4x_{1}                                                                                      ...(i)

Now,      y^{2}=3-4x

Differentiate with respect to x

\begin{array}{ll} \Rightarrow \quad & 2 y \frac{d y}{d x}=-4 \\ \therefore & \frac{d y}{d x}=\frac{-2}{y} \end{array}

             Slope of tangent =\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{-2}{y_{1}}

Here,     slope of tangent=slope of the line

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

From equation (i), we get

\begin{aligned} & 1=3-4 x_{1} \\ \Rightarrow & 4 x_{1}=2 \\ \Rightarrow & x_{1}=\frac{1}{2} \end{aligned}

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

 

 

 

 

Posted by

infoexpert23

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