#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 11 Maths Textbook Solution.

ANSWER: Equation of tangent, $x+y t^{2}=2 c t$

Equation  of normal , $x t^{3}-y t=c t^{4}-c$

HINTS:

Differentiating the given curve with respect to x and find its slope.

GIVEN:

$x y=c^{2} \text { at }\left(c t, \frac{c}{t}\right)$

SOLUTION:

\begin{aligned} &x\frac{d y}{d x}+y=0 \\ &\Rightarrow \frac{d y}{d x}=-\frac{y}{x} \end{aligned}

Slope of tangent,

$m=\left(\frac{d y}{d x}\right)\left(\operatorname{ct}, \frac{c}{t}\right)=\frac{-\frac{c}{t}}{c t}=-\frac{1}{t^{2}}$

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{c}{t}=-\frac{1}{t^{2}}(x-c t) \\ &\Rightarrow \frac{y t-c}{t}=-\frac{1}{t^{2}}(x-c t) \\ &\Rightarrow y t^{2}-c t=-x+c t \\ &\therefore x+y t^{2}=2 c t \end{aligned}

Equation of Normal  is,

\begin{aligned} &\Rightarrow y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{c}{t}=t^{2}(x-c t) \\ &\Rightarrow y t-c=t^{3} x-c t^{4} \\ &\Rightarrow x t^{3}-y t=c t^{4}-c \end{aligned}