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  • Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 13 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 15 Tangents and Normals Exercise Very short Answers Question 13 textbook solution.

Answers (1)

Answer : Angle = 0^{o}

Hint : First, find the slope of given curves then use the formula:

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

Given :

Given curve,

\begin{aligned} &y^{2}=4 x_{\text {and }} \\ &x^{2}=2 y-3 \end{aligned}

We have to find the angle between the given curves.

Solution :

Given :

       y^{2}=4x                                                                         ...(i)

      x^{2}=2y-3                                                                 ...(ii)

On differentiating equation (i) with respect to x, we get

2y\frac{dy}{dx}=4

\Rightarrow \; \; \; \; \; \frac{dy}{dx}=\frac{2}{y}

\begin{array}{ll} \Rightarrow & m_{1}=\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2} \\ \Rightarrow \quad & m_{1}=1 \end{array}

Now, differentiating equation (ii) with respect to x, we get

2 x=2 \frac{d y}{d x}

\Rightarrow \quad \frac{d y}{d x}=x

             m_{2}=\left(\frac{d y}{d x}\right)_{(1,2)}=1

\Rightarrow \quad m_{2}=1

As we know that,

          \begin{aligned} &\quad \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\Rightarrow \quad \tan \theta=\left|\frac{1-1}{1+1}\right| \\ &\Rightarrow \quad \tan \theta=0 \\ &\Rightarrow \quad \theta=0^{\circ} \end{aligned}

Hence the required angle between the curve is 0^{o}

Posted by

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