Parametric equations - locus at midpoint

Question

Consider the parametric equations $x=-2t^2$ and $y=4t$

The normal at any point, P, cuts the x-axis at Q. Find the Cartesian equation of the locus of the midpoint, M, of PQ.

Can anyone help get me started?

Answer 1

1) Find the slope of the tangent line, which is

$$m'=\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4}{-4t}=-\frac 1t$$

2) Find the slope of the normal line, which is the negative reciprocal of the slope of the tangent line.

$$m=-\frac 1{m'}=-\frac 1{-1/t}=t$$

3) Find the equation of the normal line by putting its slope and the coordinates of the point $P(x,y)=(-2t^2,4t)$ [as the point $(x_0,y_0)$ on the line] into the point-slope form of a line.

$$y=m(x-x_0)+y_0$$ $$y=t(x-(-2t^2))+4t$$ $$y=tx+2t^3+4t$$

4) Find the $x$-intercept of that normal line by substituting $y=0$ and solving for $x$. I'll let you do the details on that one and the later steps. The answer will be an expression in $t$.

5) Find the point $Q$.

6) Find the midpoint $M$ of line segment $\overline{PQ}$. The coordinates will be expressions in $t$.

7) Use the coordinates of $M$ to find a Cartesian equation that defines the locus of $M$.

That should get you started and give you a road map to finish the problem. @hkmather802, in his deleted answer, gave a somewhat shorter method that first finds the Cartesian equation of the original curve, but my answer is more straightforward (I think).