Question

The slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 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

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

Given:
x = t² + 3t − 8 and y = 2t² − 2t − 5

$$\begin{gathered} \Rightarrow \frac{dx}{dt}=2t+3\mathrm{and}\frac{dy}{dt}=4t-2 \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4t-2}{2t+3} \\ \text{The given point is (2, -1).} \\ \text{∴ x = 2 and y = -1} \\ \mathrm{Now}, \\ {t}^2+3t-8=2\text{and}2t^2-2t-5=-1 \\ \Rightarrow t^2+3t-10=0\text{and}{t}^2-t-2=0 \\ \Rightarrow \left(t+5\right)\left(t-2\right)=0\text{and}\left(t+1\right)\left(t-2\right)=0 \\ \Rightarrow t=-5,2\text{and}t=-1,2 \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}t=2\text{satisfies both of these.} \\ \therefore \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{t=2}\text{=}\frac{8-2}{4+3}\text{=}\frac{6}{7} \end{gathered}$$
Thus, the correct option is (b).