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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 21 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 21 Maths Textbook Solution.

Answers (1)

Answer: The equation of tangent \rightarrow 3x+y=4\: \: ,y=3x-4

Hint: Differentiate with respect to x.

Given: 3x^{2}-y^{2}=8, which passes through  \left ( \frac{4}{3},0 \right )

Solution: 3x^{2}-y^{2}=8                           ....(i)

Diff w.r.t x

\begin{aligned} &6 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{3 x}{y} \\ &\Rightarrow\left[\frac{d y}{d x}\right]_{\left(x_{i}, y_{i}\right)}=\frac{3 x_{1}}{y_{1}} \end{aligned}

\therefore The eq. of tangent at \left ( x_{1},y_{1} \right )is

y-y_{1}=\frac{3x_{1}}{y_{1}}\left ( x-x_{1} \right )

Tangent passes through the point \left [ \frac{4}{3},0 \right ]

\begin{aligned} &\therefore 0-y_{1}=\frac{3 x_{1}}{y_{1}}\left[\frac{4}{3}-x_{1}\right] \\ &\qquad \begin{array}{l} -y_{1}^{2}=3 x_{1}\left[\frac{4}{3}-x_1\right] \Rightarrow-y_{1}^{2}=4x_{1}-3 x_{1}^{2} \\ \text { Or } \end{array} \\ &\text { Using egn (i) } 8-3 x_{1}^{2}=4 x_{1}-3 x_{1}^{2} \end{aligned}

or\: x_{1}=2

\begin{aligned} &\therefore 3\left(2^{2}\right)-y_{1}^{2}=8 \\ &12-y_{1}^{2}=8 \Rightarrow y_{1}^{2}=4 \\ &y_{1}^{2}=\pm 2 \end{aligned}

\thereforethe points are (2,2) and (2,-2)

\thereforeThe eq. of tangent at (2,2) is :

y-2=3\left ( x-2 \right )

y=3x-4

The eq.of tangent at (2,-2) is :

y+2=-3\left ( x-2 \right )

y=-3x+4

3x+y=4

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