Question

The slope of the tangent to the curve $ x = t^2 + 3t -8 , y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is

• A. $\frac{22}{7}$
• B. $\frac{6}{7}$
• C. $\frac{7}{6}$
• D. $\frac{-6}{7}$

Answer 1

Answer: (B)

We have, $x=t^{2}+3 t-8$ and $y=2 t^{2}-2 t-5$
Put $x=2$
$\therefore 2=t^{2}+3 t-8$
$\Rightarrow t^{2}+3 t-10=0$
$\Rightarrow (t+5)(t-2)=0$
$\therefore t=-5,2$
Now, put $y=-1$
$\therefore -1=2 t^{2}-2 t-5$
$\Rightarrow 2 t^{2}-2 t-4=0$
$\Rightarrow t^{2}-t-2=0$
$\Rightarrow (t-2)(t+1)=0$
$\Rightarrow t=2,-1$
$\therefore t=2 \,\,[\because t \neq-1]$
Now, $\frac{d x}{d t}=2 t+3$ and $\frac{d y}{d t}=4 t-2$
$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}$
$=\frac{4 t-2}{2 t+3}$
$ \therefore \left(\frac{d y}{d x}\right)_{t=2}=\frac{4(2)-2}{2(2)+3}$
$=\frac{8-2}{4+3}=\frac{6}{7}$
Hence, the slope of tangent is $\frac{6}{7}$.