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

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  $\left(\frac{d y}{d x}\right)_{t=0}=\frac{0}{2}=0$