#### Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Questions Question 12 maths Textbook Solution.

$\left ( b \right )\frac{6}{7}$

Hint:

Use differentiation

Given:

$x=t^{2}+3t-8$

$y=2t^{2}+2t-5$

Solution:

Given curve are $x=t^{2}+3t-8$            (1)

And $y=2t^{2}+2t-5$                           (2)

At $\left ( 2,-1 \right )$

From (1) $t^{2}+3t-10=0$

$\Rightarrow t=2 \; or\: t=-5$

From (2) $2t^{2}-2t-4=0$

$\Rightarrow t^{2}-t-2=0$

$\Rightarrow t=2 \: \: or \: \: t=-1$

From both the solution, we get t=2

Differentiating both the equation w.r.t t, we get

$\frac{dx}{dt}=2t+3$                                                    (3)

$\frac{dy}{dt}=4t-2$                                                        (4)

Now,$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

From (3) and (4) we get

$\frac{4t-2}{2t+3}$

$\therefore \frac{dy}{dx}=\frac{4t-2}{2t+3}$     Is the slope of the tangent to the given curve.

$\left|\frac{d y}{d x}\right|_{(2,-1)}=\left|\frac{4 t-2}{2 t+3}\right|_{t-2}=\frac{8-2}{4+3}=\frac{6}{7}$         Is the slope of the tangent to the given curve at (2,-1)