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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.

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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


\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}

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