How to find the highest point on a lemniscate?

Question

A Lemniscate can be described by the following equation :

\begin{equation} \left(x^{2}+b y^{2}\right)^{2}=2 a^{2}\left(x^{2}-b y^{2}\right) \label{equa} \end{equation} where $a,b$ are parameters. (See picture below for the shape)

To find the largest $x$ that is on the curve, we can see that it will be attained in $y=0$, therefore to find $x_\text{max}$ I can simply set $y=0$ in order to find $x_\text{max}$.

How can we find $y_\text{max}$ such that $y_\text{max}$ is the largest value for $y$ that obeys the equation of this curve? Or in simpler terms, how do I compute its half-height?

Any advice or remark is always appreciated, thanks!

Answer 1

$$(x^2+by^2)^2=2a^2(x^2-by^2)\tag{1}$$

$$2(x^2+by^2)(2x+2byy')=2a^2(2x-2byy')$$

At the point of $y_{max}$ the first derivative is zero so:

$$2(x^2+by^2)(2x)=2a^2(2x)$$

$$x^2+by^2=a^2$$

$$x^2=a^2-by^2\tag{2}$$

Now replace (2) into (1):

$$a^4=2a^2(a^2-2by^2)$$

$$a^2=2a^2-4by^2$$

$$4by^2=a^2$$

$$y=\frac{a}{2\sqrt b}$$

Answer 2

Hint: Compute $dy/dx$ using implicit differentiation, Then see where that is zero. I'd guess four points (or two depending on how you set it up), compute $y$ values of these, etc.