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#### Provide solution for RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 29 maths textbook solution.

Answer : (a) is the correct option

Hint :

Slope $\frac{y-y_{1}}{x-x_{1}}$

Given : $x^{2}=4 y$

Solution :

$x^{2}=4 y$                           (1)

Differentiating (1) w.r.t x

\begin{aligned} &4 \frac{d y}{d x}=2 x \\ &\frac{d y}{d x}=\frac{x}{2} \\ &-\frac{d x}{d y}=-\frac{2}{x_{1}} \end{aligned}

Slope of normal $=-\frac{2}{x_{1}}$

Since the normal passes through $(1,2)$

\begin{aligned} &\text { Slope }=\frac{y_{1}-2}{x_{1}-1} \\ &-\frac{2}{x_{1}}=\frac{\frac{x^{2}}{4}-2}{x_{1}-2} \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\text { From }(1)] \end{aligned}

$-\frac{2}{x_{1}}=\frac{x_{1}^{2}-8}{4\left(x_{1}-2\right)} \Rightarrow x_{1}^{3}-8 x_{1}=-8 x_{1}-8 \Rightarrow x_{1}=2$

Putting the value of $x_{1}= 2$ in (1) , was $y=1$

Point of contact $(2,1)$

Equation of normal

\begin{aligned} &y-1=-\frac{2}{3}(x-2) \\ &y-1=-(x-2) \\ &x+y=3 \end{aligned}