#### Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 5 Sub Question 2 Maths Textbook Solution.

ANSWER: Equation of tangent, $y-\left(\frac{4}{5}\right)=\left(\frac{13}{16}\right)\left[x-\left(\frac{2 a}{5}\right)\right]$

Equation  of normal , $y-\left(\frac{4}{5}\right)=\left(\frac{-16}{13}\right)\left[x-\left(\frac{2 a}{5}\right)\right]$

HINTS:

Differentiate the given equation   with respect to t  and to get the slopes.

GIVEN:

$x=\frac{2 a t^{2}}{1+t^{2}}, y=\frac{2 a t^{2}}{1+t^{2}} \text { at } t=\frac{1}{2}$

SOLUTION:

Upon differentiation,

\begin{aligned} &\frac{d x}{d t}=\frac{\left(1+t^{2}\right) y a t-2 a t^{2}(2 t)}{\left(1+t^{2}\right)^{2}} \Rightarrow \frac{d x}{d t}=\frac{4 a t}{\left(1+t^{2}\right)^{2}} \\ &\Rightarrow \frac{d y}{d t}=\frac{\left(1+t^{2}\right) 6 a t^{2}-2 a t^{2}(2 t)}{\left(1+t^{2}\right)^{2}} \Rightarrow \frac{d x}{d t}=\frac{6 a t^{2}+4 a t^{4}}{\left(1+t^{2}\right)^{2}} \\ &\therefore \frac{d y}{d x}=\frac{\left(6 a t^{2}+2 a t^{4}\right)}{4 a t} \end{aligned}

$m(\text { tangent }) \text { at } t=\frac{1}{2} \text { is } \frac{13}{16}$

The normal is perpendicular to tangent, therefore , $m_{1} m_{2}=-1$

$m(\text { normal }) \text { at } t=\left(\frac{1}{2}\right) \text { is }-\frac{16}{13}$

The equation of tangent is given by,

\begin{aligned} &y-y_{1}=m(\text { tangent })\left[x-x_{1}\right] \\ &\Rightarrow y-\left[\frac{4}{5}\right]=\left(\frac{13}{16}\right)\left[x-\left(\frac{2 a}{5}\right)\right] \end{aligned}

The equation of Normal  is given by  ,

\begin{aligned} &y-y_{1}=m(\text { normal })\left[x-x_{1}\right] \\ &\Rightarrow y-\left[\frac{4}{5}\right]=\left(\frac{-16}{13}\right)\left[x-\left(\frac{2 a}{5}\right)\right] \end{aligned}