Question

Choose the correct answer in the following questions :
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

Given, $x=t^2+3t-8andy=2t^2-2t-5$ ...(i)
On differentiating w.r.t,t we get
$\therefore \frac{dx}{dt}=2t+3and\frac{dy}{dt}=4t-2\therefore \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{4t-2}{2t+3}$
The given point is (2, -1).
At x = 2, from Eq. (i), we get
$2=t^2+3t-8\Rightarrow t^2+3t-10=0$
$\Rightarrow (t-5)(t-2)=0\Rightarrow t=2ort=-5$
At y = - 1, from Eq. (i), we get
$-1=2t^2-2t-5\Rightarrow 2t^2-2t-4=0\Rightarrow (t-2)(t+1)=0\Rightarrow t=2ort=-1$
The common value of t is 2.
Hence, slope of tangent to the given curve at the given point (2,-1) is $(\frac{dy}{dx}{)}_{t=2}=\frac{4\times 2-2}{2\times 2+3}=\frac{8-2}{4+3}=\frac{6}{7}$
Hence, the correct option is (b).