The slope \(\left(\displaystyle\frac{dy}{dx}\right)\) of the tangent to the curve \(x=t^2+3t-8\), \(y=2t^2-2t-5\) at the point \((2, -1)\), is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{6}{7}\)
The point (2, −1) corresponds to the value of t that gives both coordinates.
From x = 2:
\(t^2+3t-8=2\)
\(t^2+3t-10=0\)
\((t+5)(t-2)=0,\quad t=2 \text{ or } -5\)
From y = −1:
\(2t^2-2t-5=-1\)
\(t^2-t-2=0\)
\((t-2)(t+1)=0,\quad t=2 \text{ or } -1\)
The common value is t = 2. For a parametric curve:
\( \displaystyle \frac{dx}{dt} = 2t+3,\quad \frac{dy}{dt} = 4t-2 \)
\( \displaystyle \frac{dy}{dx} = \frac{4t-2}{2t+3} \)
At t = 2:
\( \displaystyle \frac{dy}{dx} = \frac{8-2}{4+3} = \frac{6}{7} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{22}{7}\) — Incorrect. This is the slope at t = −5, which satisfies x = 2 but gives y = 50 + 10 − 5 = 55, not −1. That point is not (2, −1).
Option (3) \(\displaystyle\frac{7}{6}\) — Incorrect. This is dx/dy at t = 2, the reciprocal of the slope.
Option (4) \(\displaystyle\frac{-6}{7}\) — Incorrect. At t = 2 both dy/dt = 6 and dx/dt = 7 are positive, so the slope is positive.