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  • Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 13 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 13 maths Textbook Solution.

Answers (1)

Answer:

            \left ( d \right )\left ( 1,2 \right ) And \left ( 1,-2 \right )

Hint:

        Use differentiation

Given:

            x^{2}+y^{2}-2x-3=0

Solution:

x^{2}+y^{2}-2x-3=0

Differentiate w.r.t x, we get

\begin{aligned} &2 x+2 y \frac{d y}{d x}-2=0 \\ &2 y \frac{d y}{d x}=2-2 x \\ &\frac{d y}{d x}=\frac{2(1-x)}{2 y} \\ &\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}                                        (1)

If line is parallel to x-axis, angle with x-axis =\theta =0

Slope of x-axis =\tan \theta =\tan 0^{o}=0

Slope of tangent =Slope of x-axis

\frac{dy}{dx}=0

\frac{1-x}{y}=0

x=1

Find y When x=1

\begin{aligned} &x^{2}+y^{2}-2 x-3=0 \\ &(1)^{2}+(y)^{2}-2(1)-3=0 \\ &1+y^{2}-2-3=0 \\ &y=\pm 2 \end{aligned}

Hence, the points are \left ( 1,2 \right ) and \left ( 1,-2 \right )

               

 

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