#### Please solve RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Very Short Answer Question 2 maths textbook solution.

Answer : Slope of tangent at $t=2 \text { is } \frac{6}{7}$

Hint : Use equation of tangent,

$\frac{y-y_{1}}{x-x_{1}}=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}$

Given :

Here, $x=t^{2}+3 t-8 \text { and } y=2 t^{2}-2 t-5$

We have to find the slope of tangent to given curve at $t = 2$

Solution :

The equation of the given curve is

\begin{aligned} &x=t^{2}+3 t-8 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(i)\\ &y=2 t^{2}-2 t-5 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii) \end{aligned}

Differentiating equation (i) with respect to $'t'$

We have,

\begin{aligned} &\frac{d y}{d t}=4 t-2\\ &\therefore \frac{d y}{d x}=\frac{\frac{d x}{d t}}{\frac{d y}{d t}}=\frac{4 t-2}{2 t+3} \end{aligned}

Now, slope of tangent at $t=2$

$\left(\frac{d y}{d x}\right)_{t=2}=\frac{8-2}{4+3}=\frac{6}{7}$

Hence, Slope of tangent at $t=2$ is $\frac{6}{7}$.