Can we show the locus of the midpoint of $PQ$ is an ellipse?

Can we show the locus of the midpoint of $PQ$ is an ellipse?

Question

The points $P$ and $Q$ on the ellipse

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

have coordinates $(a\cos\theta, b\sin \theta)$ and $(-a\sin\theta, b\cos\theta)$ respectively. Show that, if $O$ is the origin,

1. $OP^2+OQ^2 = a^2 + b^2$,

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1. the area of triangle $OPQ$ is $\dfrac{1}{2}ab$,

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1. the midpoint of $PQ$ always lies on the curve whose equation is

$$\dfrac{2x^2}{a^2}+\dfrac{2y^2}{b^2}=1.$$

Suggestion

Could we start by sketching the graph and marking the points $P$ and $Q$?

When finding the area of the triangle $OPQ$, are there any other areas we could find more easily that could help us?

Solution

Let $c = \cos \theta$ and $s = \sin \theta$, and so $c^2 + s^2 = 1.$

Thus $OP^2+OQ^2 = (ac-0)^2 + (bs-0)^2 + (-as-0)^2 + (bc-0)^2 =a^2(c^2+s^2) + b^2(s^2+c^2) = a^2 + b^2$.

The area of triangle $OPQ$ is the area of the rectangle in the diagram with three right-angled triangles subtracted.

So the area of triangle $OPQ$ is

$$(ac+as)(bs)-(0.5ac\times bs)-(0.5as\times bc) -0.5(ac+as)(bs-bc)$$ $$= abcs+abs^2-0.5abcs-0.5abcs -0.5abcs+0.5abcs -0.5abs^2 +0.5abc^2$$ $$= \dfrac{ab}{2}.$$

We have here chosen the case when $b\sin \theta > b\cos\theta$, but this method will deal with all possible cases.

The midpoint of $PQ$ is the point $(\dfrac{ac-as}{2}, \dfrac{bc+bs}{2}) =(x,y).$

Thus $\dfrac{2x}{a}= c-s, \dfrac{2y}{b} = c+s.$ Squaring each equation and then adding them gives

$$\dfrac{4x^2}{a^2}+\dfrac{4y^2}{b^2} =2,$$

and so the midpoint of $PQ$ has the locus

$$\dfrac{2x^2}{a^2}+\dfrac{2y^2}{b^2}=1.$$

We can see the locus traced out by $A$ in the GeoGebra applet below.

Download GeoGebra file