Is this a True Property of the Lemniscate of Bernoulli?

Question

I am trying to figure out if the following is true:

Take the Lemniscate of Bernoulli (a plane curve defined from two given points F1 and F2, known as foci, at distance 2a from each other as the locus of points P so that PF1·PF2 = a^2).

Now imagine plotting it on the x,y axis.

Now imagine a straight line starting at either of the focal points and traveling in a direction perpendicular to the x-axis until it collides with P (the path of the lemniscate).

Is it true that the length of this line = a/2? And if so, why?

This is an illustration of what I'm asking (ignore right-hand side):

I'm doing some graphic design that depends on this being true, and it seems to be true visually, but I'm not a mathematician and it would be really useful to know if it is a guaranteed fact for some reason.

Thankyou for any help.

Answer 1

The equation for this curve in cartesian coordinates is given by

$$(x^2 + y^2)^2 = 2a^2 (x^2 - y^2).$$

We now just need to chck whether a vertical line through one of the focii intersectes this curve at a point of distance $a/2$ to the corresponding focus. Using the right focal point, such a line is given by $0 = x-a$. Now we can plug this into the original equation and solve for the $y$ coordinate of the intersection points:

$$(a^2+y^2)^2 = 2a^2(a^2-y^2).$$

After bringing both terms to the same side and simplifying, we get the equation

$$-a^4 + 4 a^2 y^2 + y^4 = 0$$

We can easily solve this equation as it is a quadratic equation in $u=y^2$ and get the (real) solutions

$$y = \pm \sqrt{\sqrt{5} - 2} \cdot a$$

You conjectured $y = \pm 0.5 a$ but if we compute this term we see that this is not true, our solution shows that we have $y = \pm 0.485868.... a$, which is quite close.

Answer 2

It's true that the Lemniscate's highest points are $a/2$ units above the horizontal line of symmetry. However, these points are not directly above the foci.

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Consider the Cartesian equation: $$(x^2+y^2)^2=2a^2(x^2-y^2) \tag{1}$$

Substituting $x=a$ and $y=a/2$ gives $$\frac{a^4}{16} = 0 \tag{2}$$ This equation isn't valid (unless $a$ itself is zero, a case we reasonably ignore), so the points $a/2$ units above the foci are not on the curve. (As @flawr's answer indicates, they're quite close.)

On the other hand, substituting $y=a/2$ and leaving $x$ as-is gives $$\frac1{16} \left(3 a^2 - 4 x^2\right)^2=0 \tag{3}$$ This equation has exactly two solutions, $x = \pm a \sqrt{3}/2$, which indicates that the line $y=a/2$ meets the curve at exactly two points; clearly, these must be the highest points on the curve. (A slightly-lower line would meet the curve in four points; a slightly-higher one would meet the curve not-at-all.)

Answer 3

Use the pythagorean theorem to find $PF2$ in terms of $PF1$ and $a$ and use the identity $PF1*PF2=a^2$ to get biquadratic equation for $PF2$.