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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 9 Maths Textbook Solution.

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ANSWER: Equation of tangent, m^{2} x-m y+a=0

                 Equation  of normal , m^{2} x+m^{3} y-2 a m^{2}-a=0


 Differentiating the given curve and find its slope.


y^{2}=4 a x \text { at }\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)


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

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{2 a}{m}=m\left(x-\frac{a}{m^{2}}\right) \\ &\Rightarrow \frac{m y-2 a}{m}=m \frac{m^{2} x-a}{m^{2}} \\ &\Rightarrow m y-2 a=m^{2} x-a \\ &\Rightarrow m^{2} x-m y+a=0 \end{aligned}

Equation of Normal  is,

\begin{aligned} &\Rightarrow y-y_{1}=\frac{-1}{\text { slope of tangent }}\left(x-x_{1}\right) \\ &\Rightarrow y-\frac{2 a}{m}=-\frac{1}{m}\left(x-\frac{a}{m^{2}}\right) \\ &\Rightarrow \frac{m y-2 a}{m}=-\frac{1}{m} \frac{m^{2} x-a}{m^{2}} \\ &\Rightarrow m^{3} y-2 a m^{2}=-m^{2} x+a \\ &\Rightarrow m^{2} x+m^{3} y-2 a m^{2}-a=0 \end{aligned}

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