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  • Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question, Question 22 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 22 maths textbook solution. 

Answers (1)

Answer : (c) is the correct option.

Hint :

\tan \theta=\left|\frac{m_{1}-m_{2}}{1-m_{1} m_{2}}\right|

Given :   y^{2}=4 a x \text { And } x^{2}=4 a y

Solution :

\begin{aligned} &y^{2}=4 a x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (1)\\ &x^{2}=4 a y\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (2) \end{aligned}

Differentiating (1) w.r.t x we get

\begin{aligned} &2 y \frac{d y}{d x}=4 a \\ &\left|\frac{d y}{d x}\right|_{(0,0)}=\frac{4 a}{2 x}=0=m_{1} \end{aligned}

And

\begin{aligned} &2 x=4 a \frac{d y}{d x} \\ &\left|\frac{d y}{d x}\right|_{(0,0)}=\left|\frac{4 a}{2 x}\right|=0=m_{2} \end{aligned}

\begin{aligned} &\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|= \\ &=\left|\frac{0-0}{1+0 \cdot 0}\right|=\infty \\ &\theta=\frac{\pi}{2} \end{aligned}

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