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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.

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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 )


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

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



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