Focus and directrix of a quadratic curve

Question

I have been trying to solve the following problem for a while.

Consider the quadratic curve $(1-e^2)x^2+y^2-2c(1+e)x=0$, where $0<e<1$ or $e>1$. Also, consider the point $F'=(\frac{c(1+e)}{1-e},0)$ and the line $x=\frac{c(1+e)}{e(1-e)}$. Prove that $F'$ is the second focus and $x$ is the directrix of the quadratic curve.

I know that if $e>1$, the curve is a hyperbola. And that if $0<e<1$, the curve is an ellipse. But I haven't been able to establish a proper way to solve this problem. Any help is appreciated.

Answer 1

What you have to prove is that for any point $M$ of the conic curve, one has:

$$eMH=MF'\tag{1}$$

where $H$ is the orthogonal projection of $M$ onto the (vertical) directrix.

(1) is equivalent to its squared version $e^2MH^2=MF'^2$ (because we deal with lengths), i.e., analyticaly:

$$e^2\left(x-\dfrac{c(1+e)}{e(1-e)}\right)^2=\left(x-\dfrac{c(1+e)}{1-e}\right)^2+(y-0)^2.$$

Reducing to the same denominator, and keeping only the numerators:

$$(e(1-e)x-c(1+e))^2-((x(1-e)-c(1+e))^2+(1-e)^2y^2=0$$

which is equivalent to:

$$(e^4 - 2e^3 + 2e - 1)x^2 + 2c(e^3 - e^2 - e + 1)x - (e^2-2e+1)y^2=0.$$

The 3 coefficients can be divided by $(e-1)^2 \ne 0$, giving back the initial equation:

$$(1-e^2)x^2-2c(1+e)x+y^2=0$$

As we have used equivalences all the way long, the fact is proven.

Answer 2

For $1>e>0$

$$\left(\left(1-e^{2}\right)x^{2}+2c\left(1+e\right)x+\frac{4c^{2}\left(1+e\right)}{\left(1-e\right)}\right)+y^{2}=\frac{4c^{2}\left(1+e\right)}{\left(1-e\right)}$$

$$\left(\sqrt{\left(1-e^{2}\right)}x+\sqrt{\frac{4c^{2}\left(1+e\right)}{\left(1-e\right)}}\right)^{2}+y^{2}=\frac{4c^{2}\left(1+e\right)}{\left(1-e\right)}$$

Shift origin to $\left(-\frac{2c}{1-e},0\right)$

$$(1-e^2)X^{2}+y^{2}=\frac{4c^{2}\left(1+e\right)}{\left(1-e\right)}$$

Noe you can solve for the details of the ellipse quite easily. Similarly you can handle the case of $e>1$