Question

The slope of the tangent to the curve x = t² + 3 t − 8, y = 2t² − 2t − 5 at point (2, −1) is

(a) 22/7
(b) 6/7
(c) −6
(d) none of these

Answer 1

(b) 6/7

$$\begin{gathered} x=t^2+3t-8\mathrm{and}y=2t^2-2t-5 \\ \frac{dx}{dt}=2t+3\mathrm{and}\frac{dy}{dt}=4t-2 \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{4t-2}{2t+3} \\ \mathrm{The}\mathrm{g}\text{iven point is (2, -1).} \\ \text{∴}\mathit{\text{x}}\text{=2 and}\mathit{\text{y}}\text{=-1} \\ \mathrm{Now}, \\ {t}^2+3t-8=2\text{and}2t^2-2t-5=-1 \\ \text{Let us solve one of these to get the value of}\mathit{\text{t.}} \\ {t}^{\mathit{2}}+3t-10\mathit{=}\mathit{0}\mathit{}\mathrm{and}2t^{\mathit{2}}\mathit{-}2t-4\mathit{=}\mathit{0} \\ \Rightarrow \left(t+5\right)\left(t-2\right)=0\mathit{}\mathrm{and}\mathit{}\left(2t+2\right)\left(t\mathit{-}2\right)\mathit{=}\mathit{0} \\ \Rightarrow t=-5\mathit{}\text{or}\mathit{\text{t=}}\text{2 and}\mathit{\text{t}}\text{=-1 or}\mathit{\text{t}}\text{=2} \\ \mathrm{These}\mathrm{two}\mathrm{have}\mathit{}t=2\mathrm{as}\mathrm{a}\mathrm{common}\mathrm{solution}. \\ \therefore \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{t=2}\text{=}\frac{8-2}{4+3}=\frac{6}{7} \end{gathered}$$