Question

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

• A. 7/6
• B. 2/3
• C. 3/2
• D. 6/7

Answer 1

The correct option is D 6/7

We first determine the value of $t$ corresponding to the given values ofx and $y$. From $t^2+3t-8=2$, we get $t=2,-5$, and from $2t^2-2t-5=2$ we get $t=2,-1$. Hence to the given point there corresponds the value $t=2$. Therefore, the slope of the tangent at $(2,-1)$ is
${\left|y^{\prime }\right|}_{t=2}={\left|\frac{dy/dt}{dx/dt}\right|}_{t=2}={\left|\frac{4t-2}{2t+3}\right|}_{t=2}=\frac{6}{7}$