#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 18 Maths Textbook Solution.

ANSWER: Equation of tangent, $y y_{1}=2 a\left(x+x_{1}\right)$

Equation  of normal , $y-y_{1}=\frac{-y_{1}}{2 a}\left(x-x_{1}\right)$

HINTS:

Differentiating  with respect to x  and find its slope.

GIVEN:

$y^{2}=4 a x \text { at }\left(x_{1}, y_{1}\right)$

SOLUTION:

\begin{aligned} &2 y \frac{d y}{d x}=4 a \\ &\Rightarrow \frac{d y}{d x}=\frac{2 a}{y} \end{aligned}

$\text { Slope of tangent }=\frac{d y}{d x_{\left.\mid x_{1}, y_{1}\right\}}}=\frac{2 a}{y_{1}}=m$

Equation of tangent is

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-y_{1}=\frac{2 a\left(x-x_{1}\right)}{y_{1}} \end{aligned}

\begin{aligned} &\Rightarrow y y_{1}-y_{1}^{2}=2 a x-2 a x_{1} \\ &\Rightarrow y y_{1}-4 a x_{1}=2 a x-2 a x_{1} \\ &\Rightarrow y y_{1}=2 a x+2 a x_{1} \\ &\therefore y y_1=2 a\left(x+x_{1}\right) \end{aligned}

Equation of Normal  is ,

$y-y_{1}=-\frac{1}{\text { slope of tangent }}\left(x-x_{1}\right)$

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

$\Rightarrow y-y_{1}=\frac{-y_{1}}{2 a}\left(x-x_{1}\right)$