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 $(x,y)$?
Best Answer
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.