find largest $x$ based on $f(x,y)=c$

Question

Consider the following equation: $$ \frac{p}{\left(a-x\right)^2+y^2}+\frac{1-p}{\left(b-x\right)^2+y^2}=1 $$

where $0<p<1$ and $a,b\in\mathbb{R}$.

I want to look for the largest $x$ satisfying the equation. This should be a basic calculus question, finding the largest $x$ such that $f(x,y)=1$. However I do not know where to start?

Thanks a lot!

Edit: I also tried using Lagrange multipliers, maximising the function $f(x)=x$ under the constraint of above. However I do not get solutions in $\mathbb{R}$... Especially in the example under (with $p=0.25, a=1,b=0$).

$p=0.25, a=1,b=0$">

Answer 1

The question makes sense only if $0<p<1$. In that case $x$ attains its extrema at those points where $dx/dy=0$. Differentiating the given equation one finds: $$ {dx\over dy}=y\ \frac{{p\left((b-x)^2+y^2\right)^2}{}+ {(1-p)\left((a-x)^2+y^2\right)^2}{}} {{p(a-x)\left((b-x)^2+y^2\right)^2}{} +{(1-p)(b-x)\left((a-x)^2+y^2\right)^2}{}} $$ and this vanishes only for $y=0$. Hence to find the extrema of $x$ you have to solve the equation $$ \frac{p}{\left(a-x\right)^2}+\frac{1-p}{\left(b-x\right)^2}=1. $$ Unfortunately this is a fourth degree equation, not so easy to solve explicitly (if that is what you want). The equation can have two real and two complex solutions, in which case the real solutions give the extrema of $x$. Or it can have four real solutions, in which case the two intermediate values must be discarded.

For example, for $p=0.25$, $a=1$ and $b=0$ one gets only two real solutions: $x_1\approx-0.897745$ and $x_2\approx1.59536$, which are the minimum and maximum value of $x$.

Answer 2

If you look at a sketch of the function, you see that it is symmetrical about the $x$ axis. So it should reduce to an equation in $y^2$ (really obvious if you look at the equation...). At the extrema $y = 0$. If you take it as an equation for $y$, the points you are looking for are double zeros (for different $x$, there are two distinct $y$, for larger $x$ there are no -- real -- $y$, thus $y^2$ becomes negative).