Finding foot of perpendicular of centre of ellipse on its variable tangent 
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 $π/2.$
Ref:Locus of foot of perpendicular of origin from tangent of ellipse, related

The points may be given as $(a\cos \alpha, b\sin \alpha)$ and $(-a\sin \alpha , b \cos \alpha )$, then their midpoint is given as: $\left( \frac{a}{2} (\cos \alpha - \sin \alpha) , \frac{b}{2} (\cos \alpha + \sin \alpha) \right)=(\kappa, \beta)$, the equation of chord given midpoint of ellipse is:
$$ \frac{\kappa X}{a^2} + \frac{\beta Y}{b^2} = \frac{\kappa^2}{a^2} + \frac{\beta^2}{b^2} $$
Let $u= \frac{\cos \alpha - \sin \alpha}{2}$ and $ v=\frac{\cos \alpha + \sin \alpha}{2}$, then our equation becomes:
$$\frac{ uX}{a} + \frac{ vY}{b} = u^2 + v^2 $$
Foot of perpendicular from origin for above line is given as:
$$ \frac{ax}{u}= \frac{by}{v}= \frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
$ax= u(\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$
and $by =v (\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$, following the last line of this answer, I squared and added:
$$(ax)^2 + (by)^2 = (ab)^4  \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2= \frac{(ab)^4}{4} \left[ \frac{1}{b^2 u^2 + a^2 v^2}\right]^2$$
How do I write bracketed term in purely $(x,y)$?
 A: Here's one method to simplify things. First note that :
$$u^2+v^2=1/2$$
Now, consider the equation you found:
$$\frac{ax}{u}= \frac{by}{v}= \frac{a^2b^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
Then:
$$axv=byu\Rightarrow a^2v^2=b^2u^2y^2/x^2\Rightarrow b^2u^2+a^2v^2=b^2u^2\left(\frac{x^2+y^2}{x^2}\right)$$
Plugging this back into one of the equations gives:
$$\frac{ax}{u}=\frac{a^2b^2x^2}{2b^2u^2(x^2+y^2)}\Rightarrow \boxed{u=\frac{ax}{2(x^2+y^2)}}$$
Similarly you will also find
$$\boxed{v=\frac{by}{2(x^2+y^2)}}$$
Squaring and adding them, along with the first equation at the extreme top gives:
$$\frac12=u^2+v^2=\frac{a^2x^2}{4(x^2+y^2)^2}+\frac{b^2y^2}{4(x^2+y^2)^2}$$
$$\Rightarrow \boxed{2(x^2+y^2)^2=a^2x^2+b^2y^2}$$
A: You are missing a $(u^2+v^2)$ when you add $(ax)^2$ and $(by)^2$.
Just in case, you want to finish continuing from the point you got to,
$(ax)^2 + (by)^2 = (ab)^4 (u^2+v^2)  \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2$
$ = \cfrac{(ab)^4}{2}  \left[ \frac{(u/v)^2 +1}{b^2 (u/v)^2 + a^2} \right]^2$
Now note that $\cfrac{u}{v} = \cfrac{ax}{by}$
That leads to $2 (x^2+y^2)^2 = (ax)^2 + (by)^2$
or we could have substituted $u = k a x, v = k by$ in RHS.
