Bernoulli's Lemniscate: Extremes and Intersection with the axis

Question

Given the equation:

$$(x^2+y^2)^2 - 2p^2(x^2 + y^2) = k^4 - p^4$$

Determine the extremes and intersections with the axis.

I know it requires the use of implicit functions but I'm not sure how to proceed.

Answer 1

There is a typo in your equation :$\quad (x^2+y^2)^2 - 2p^2(x^2 + y^2) = k^4 - p^4.\quad$ This is the equation of two concentric circles.

The correct equation is : $$(x^2+y^2)^2 - 2p^2(x^2 - y^2) = k^4 - p^4$$ Moreover, this is not the equation of a lemniscate. This is the equation of a family of Cassini Ovals :

The lemniscate is only the particular case of $k=p$ (drawn in red).

The equation of the lemniscate, such as drawn in your question is : $$(x^2+y^2)^2 - 2p^2(x^2 - y^2) = 0$$ The intersects with the $x$-axis is for $y=0$, thus the roots of $x^4 - 2p^2x^2 = 0$,

i.e. the points $(-\sqrt{2}\,p,0)$ , $(0,0)$ and $(\sqrt{2}\,p,0)$.

The extremes of the lemniscate are located for $\frac{dy}{dx}=0$. Differentiating the equation of the lemniscate leads to : $2(x^2+y^2)(2xdx+2ydy)-2p^2(2xdx-2ydy)=0$.

$2(x^2+y^2)(2xdx)-2p^2(2xdx)=0$ gives $x^2+y^2=2p^2$. Putting it into the equation of the lemniscate leads to : $p^4-2p^2(2x^2-p^2)=0$ , then $x=\pm\frac{\sqrt{3}}{2}p$ and $y=\pm\frac12 p$.

The extremes are : $(\frac{\sqrt{3}}{2}p\:,\:\frac12 p)$ , $(-\frac{\sqrt{3}}{2}p\:,\:\frac12 p)$ , $(\frac{\sqrt{3}}{2}p\:,\:-\frac12 p)$ and $(-\frac{\sqrt{3}}{2}p\:,\:-\frac12 p)$.