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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.

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

Answers (1)

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}

 

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