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 $\frac{6}{k}$. The value of $k$ is ____.

Answer 1

7

Answer 2

To find the slope of the tangent to the curve defined parametrically by $x=t^2+3t-8$ and $y=2t^2-2t-5$, we first find the derivatives of $x$ and $y$ with respect to the parameter $t$: $\frac{dx}{dt} = \frac{d}{dt}(t^2+3t-8) = 2t+3$ $\frac{dy}{dt} = \frac{d}{dt}(2t^2-2t-5) = 4t-2$ The slope of the tangent, $\frac{dy}{dx}$, is given by the ratio $\frac{dy/dt}{dx/dt}$: $\frac{dy}{dx} = \frac{4t-2}{2t+3}$ Next, we need to find the value of the parameter $t$ that corresponds to the given point $M(2, -1)$. For the x-coordinate: $t^2+3t-8 = 2 \Rightarrow t^2+3t-10 = 0 \Rightarrow (t-2)(t+5) = 0$ This gives $t=2$ or $t=-5$. For the y-coordinate: $2t^2-2t-5 = -1 \Rightarrow 2t^2-2t-4 = 0 \Rightarrow t^2-t-2 = 0 \Rightarrow (t-2)(t+1) = 0$ This gives $t=2$ or $t=-1$. The common value of $t$ for the point $(2, -1)$ is $t=2$. Now, we substitute $t=2$ into the expression for $\frac{dy}{dx}$ to find the slope of the tangent at point $M$: $\frac{dy}{dx}\bigg|_{t=2} = \frac{4(2)-2}{2(2)+3} = \frac{8-2}{4+3} = \frac{6}{7}$ The problem states that the slope of the tangent is $\frac{6}{k}$. Therefore, we equate our calculated slope to the given form: $\frac{6}{k} = \frac{6}{7}$ Solving for $k$, we get: $k = 7$