$\frac{x^2}{a^2+ \lambda}+ \frac{y^2}{b^2+\lambda}=1$ family with $-a^2< \lambda < -b^2$ orthogonal to family with $\lambda > -b^2>-a^2$

Question

Here is the problem:

Prove that the family $\frac{x^2}{a^2+ \lambda}+ \frac{y^2}{b^2+\lambda}=1$ with $-a^2< \lambda < -b^2$ is orthogonal to the with family $\lambda > -b^2>-a^2$.

In particular I seek a solution via complex analysis ( so no geometric arguments etc.).

Therefore, I assumme we will want to make use of the following fact:

For $f(x,y)=u(x,y)+iv(x,y)$ analytic, the families $u(x,y)=\alpha$ and $v(x,y)= \beta$ are orthogonal.

The obvious choice of $u(x,y)$ and $v(x,y)$ doesn't seem to work as the Cauchy-Riemann equations won't be satisfied.

If anyone could help me out here that would be great. Many thanks.

EDIT: On second thought, having been reminded of the equality sign, I think the obvious choice does work. That is, let $u(x,y) = \frac{x^2}{a^2+ \lambda}+ \frac{y^2}{b^2+\lambda}=1$ and $v(x,y)=\frac{x^2}{a^2+ \kappa}+ \frac{y^2}{b^2+\kappa}=1$, where $-a^2< \lambda < -b^2$ and $\kappa > -b^2>-a^2$. Then $f(z)=u+iv=1+i$ is clearly analytic and we get the desired result. The argument is a bit dodgy, but is something in that direction true?

Answer 1

Hint: The ellipses $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1,~~ a>c ~~~(1)$$ and the hyperbolas $$\frac{x^2}{d^2}+\frac{y^2}{d^2-c^2}=1,~~ c>d ~~~~(2)$$ have common foci at $(-c,0)$ and $(c,0)$. These are both confocal and orthogonal.

One complex variable connection:

Let $w=u+iv$, the the conformal transformation leads us to $$z=g\sin w\implies x+iy=g \sin u \cosh v +i g \cos u \sin h v~~~~(3)$$ $$\implies x= \sin u \cosh v, ~~ y= \cos u \sinh v~~~~(4)$$ We can eliminate $v$ to get $$\frac{x^2}{g^2\sin^2 u}- \frac{y^2}{g^2\cos ^2 u}=1~~~~(5)$$ which is nothing but (1)

Next, by ellminating $u$ we get $$\frac{x^2}{g^2 \cosh^2 v}+ \frac{y^2}{g^2 \sinh^2 v}=1 ~~~~(6)$$ Which is nothing but (2).

The other complex variable connection is thought provoking which would mean $$w=\sin^{-1}(h z) \implies u+iv=F(x,y)+i G(x,y)$$ Then $F(x,y)=u$ and $G(x,y)=v$ will be orthogonal trajectories. I wonder how to get them. It will be interesting if some one gets them?!

Answer 2

Elliptic coordinates

$$z=c\cosh w \iff (x,y)=(c\cosh u \cos v,c\sinh u \sin v)$$

which satisfies Cauchy-Riemann relations.

Confocal elliptic coordinates

$$\frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}=1 \tag{$a > b\ge 0$}$$

How they relate

• $c^2=a^2-b^2$

• family of ellipses, $-b^2<s$

\begin{align} s &= a^2\sinh^2 u-b^2\cosh^2 u \\ 1 &= \frac{x^2}{c^2\cosh^2 u}+\frac{y^2}{c^2\sinh^2 u} \\ e &= \operatorname{sech} u \end{align}

• family of hyperbolae, $-a^2<s<-b^2$

\begin{align} s &= -a^2\sin^2 v-b^2\cos^2 v \\ 1 &= \frac{x^2}{c^2\cos^2 v}-\frac{y^2}{c^2\sin^2 v} \\ e &= |\sec v \,| \end{align}

Please also refer another answer of mine here.