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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 $(x,y)$?

tryst with freedom
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2 Answers2

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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:

$$\frac12=u^2+v^2=\frac{a^2x^2}{4(x^2+y^2)^2}+\frac{b^2y^2}{4(x^2+y^2)^2}$$ $$\Rightarrow \boxed{2(x^2+y^2)^2=a^2x^2+b^2y^2}$$

Yuzuriha Inori
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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.

Math Lover
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