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

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}

$\therefore$the points are (2,2) and (2,-2)

$\therefore$The 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 :

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

$y=-3x+4$

$3x+y=4$