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  • Provide Solution for RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Fill in the blanks Question 6

Provide Solution for RD Sharma Class 12 Chapter 15 Tangents and Normals  Exercise Fill in the blanks  Question 6

Answers (1)

Answer:

                Slope of tangent =0

Hint:

First put x= 1 in equation x= 3t^{2}+1 , then find t and  \frac{dy}{dx}.

Given:

Given curve,

             x= 3t^{2}+1   and  y= t^{3}-1

To find:

We have to find the slope of tangent to the given curve at  x= 1

Solution:

Given,

                x= 3t^{2}+1                                                                                                                                         … (i)

                y= t^{3}-1                                                                                                                                            … (ii)

Put x= 1  in equation (i), we get

\begin{aligned} & & x=3 t^{2}+1 \\ \Rightarrow & & 1=3 t^{2}+1 \\ \Rightarrow & & 3 t^{2}=0 \\ \Rightarrow & & t=0 \end{aligned}

Differentiating equation (i) and (ii) with respect to x , we get

\begin{aligned} &\Rightarrow \quad \frac{d x}{d t}=6 t \text { and } \frac{d y}{d t}=3 t^{2} \\\\ &\therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}

=\frac{3 t^{2}}{6 t}=\frac{t}{2}

Thus we get the slope of tangent  \left(\frac{d y}{d x}\right)_{t=0}=\frac{0}{2}=0

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