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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.

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:


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)





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