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. 
 A: 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.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
A: The envelop of all your chords is the ellipse itself scaled down by $\sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
