Seems familiar?

Let B be any point on the ellipse. Then let the point be $(a\cos t,b\sin t)$ .

Tangent to the ellipse at $(a\cos t,b\sin t)$ has slope $-\frac{b\cos t}{a\sin t}$ .

Then the normal at this point would have slope $\frac{a\sin t}{b\cos t}$

Equation of the normal line:

$\frac{a\sin t}{b\cos t}=\frac{y-b\sin t}{x-a\cos t}$

Writing the same thing in intercept form,

$\frac{x}{\left(\frac{\left(a^2-b^2\right)\cos t}{a}\right)}+\frac{y}{\left(-\frac{\left(a^2-b^2\right)\sin t}{b}\right)}=1$

But since this normal line passes through the end of the minor axis, it also intercepts the y-axis at $y=-b$ . Therefore,

$-\frac{\left(a^2-b^2\right)\sin t}{b}=-b$

$\implies \sin t\ =\ \frac{\left(\frac{b^2}{a^2}\right)}{\left(1-\frac{b^2}{a^2}\right)}=\frac{1-e^2}{e^2}$

Also, since the normal line cuts the latus rectum at $x=a\cos t$ , and the focus' coordinates are $(ae,0)$ , $ae=a\cos t \implies e=\cos t$

Now, eliminating the trigonometric functions:

$\left(\frac{1-e^2}{e^2}\right)^2+\left(e\right)^2=1$

$\implies e^6-2e^2+1=0$

Using the rational root theorem, we find that $e=1,-1$ are the integral roots. So $e^2 - 1$ is a factor of $e^6-2e^2+1$ . Dividing $e^6-2e^2+1$ by $e^2 - 1$ , we find out that the other factor is $e^4 + e^2 - 1$ .

So, $e^6-2e^2+1=0$ can be written as $(e^2 - 1)(e^4 + e^2 - 1)=0$ .

From here, $e=1$ or $e=-1$ . But for an ellipse, $e$ is a ratio and $e<1$ always for an ellipse. So $e=1,-1$ are not true.

What remains is $e^4 + e^2 - 1=0$ , which is true if we check for $e$ solving this biquadratic, we find one solution which can be the solution to the eccentricity of an ellipse.

So, $\boxed{e^4 + e^2 - 1=0}$ is true.