Find the equation of the locus of the mid-point between an elliptical point and its directrix

Question

I'm struggling with this question:

The point $P$ lies on the ellipse $x^2+4y^2=1$ and $N$ is the foot of the perpendicular from $P$ to the line $x=2$. Find the equation of the locus of the mid-point of $PN$ as $P$ moves on the ellipse.

It seems that the locus should be a circle, but I'm not sure how to find the actual equation for it. If $F$ is a focus, I know that the defining characteristic of an ellipse is that $0 < \frac{PN}{PS} < 1$. How do I solve this problem?

Answer 1

Hint: The line x=2 is a vertical line. If P=(x,y) is a point on the ellipse, the perpendicular to the line x=2 will always meet it at (2,y). So the mid-point is $(\frac{1}{2}(x+2), y)$.

Updated: If you're still stuck, here's the next step: use the parametrization of the ellipse $x=\cos(\theta)$, $y=\frac{1}{2}\sin(\theta)$ in the mid-point expression above. So the mid-point is $$ 1+\frac{\cos(\theta)}{2}, \frac{1}{2}\sin(\theta) $$

Now you can use $\cos^2 \theta + \sin^2 \theta= 1$ to get the equation of the locus.

Answer 2

I do not see why you need to concern yourself with the loci of the ellipse. You are given an equation in Cartesian coordinates and you are asked for another equation in Cartesian coordinates--you can just stick with equations.

For a given point $P(\pm\sqrt{1-4y^2},y)$ on the ellipse, the midpoints are clearly found as expressions in $y$. Just find an equation that covers both midpoints. This is most easily done by simplifying the equation for $x$ to get rid of the plus-or-minus.

(You do end up with the equation for a circle.)