Find the locus of the foot of perpendicular from the centre of the ellipse. Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
 on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
 Consider two points $P$ and $Q$ such that $P(a\cos\theta , b\sin\theta),\; Q(a\sin\theta, -b\cos\theta).\;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $\;(X_1,Y_1)\;$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way. 

Let the centre of the ellipse be O and the foot be R.  Also let point P is $(acos\theta, bsin\theta)$ then Q is $(acos(\theta+\frac{\pi}{2}), bsin(\theta+\frac{\pi}{2})$ or $(-asin\theta, bcos\theta)$ then PQ is
$$\frac{y - bcos\theta}{x + asin\theta} = \frac{b(sin\theta - cos\theta)}{a(sin\theta + cos\theta)} or  \frac{\frac{y}{b} - cos\theta}{\frac{x}{a} + sin\theta} = \frac{sin\theta - cos\theta}{sin\theta + cos\theta}$$
Also OR is
$$\frac{y}{x} = - \frac{a(sin\theta + cos\theta)}{b(sin\theta - cos\theta)} or  \frac{by}{ax} = -\frac{sin\theta + cos\theta}{sin\theta - cos\theta}$$
From these two equations we have
$$x^2 + y^2 = - axsin\theta + bycos\theta$$
Divide both sides by $\sqrt{a^2x^2 + b^2y^2}$ and let $tan\phi = \frac{by}{ax}$ then we have
$sin(\theta - \phi) = \frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}$ or $\theta = sin^{-1}(\frac{x^2 + y^2}{\sqrt{a^2x^2 + b^2y^2}}) + \phi$
Also from the second equation
$$\frac{y}{x} = - \frac{a}{b}\frac{tan\theta + 1}{-1 + tan\theta}$$
Substitute $\theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
