#### Provide solution for RD Sharma Class 12 Chapter 15 Tangents and Normals Exercise Multiple Choice Question , Question 26 maths textbook solution.

Answer : (b) is the correct option

Hint : $\text { Putting } \mathrm{x}=2 \text { and } \mathrm{y}=1 \text { in }(1) \text { and }(2)$

Solution :

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

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

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

Putting $x=2$ in (1)

\begin{aligned} &2=t^{2}+3 t-8 \\ &t^{2}+3 t-10=0 \\ &t^{2}+5 t-2 t-10=0 \end{aligned}

\begin{aligned} &t(t+5)-2(t+5)=0 \\ &(t+5)(t-2)=0 \\ &t=-5,2 \end{aligned}

Putting $y = -1$ in (2)

\begin{aligned} &-1=2 t^{2}-2 t-5 \\ &2 t^{2}-2 t-4=0 \\ &2 t^{2}-t-2=0 \\ &t^{2}-2 t+t-2=0 \end{aligned}

\begin{aligned} &t(t-2)+1(t-2)=0 \\ &(t-2)(t+1)=0 \\ &t=2,-1 \end{aligned}

Putiing $t =2$ in (3)

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

The slope of tangent is $\frac{6}{7}$