Statement: The equation of a conic with a focus at the origin $Ax+By+\sqrt{x^2+y^2}=c^2$ satisfies $A^2+B^2=1+2hc^2$. Show that the eccentricity and the semi-major axis of the conic are $\varepsilon=\sqrt{1+2hc^2}$ and $a=-1/(2h)$ (for $h\neq0$).
Approach: First of all, for me it's strange that a conic is given by a non-quadratic expression. So what I did was to square both sides of equation $Ax+By+-c^2=-\sqrt{x^2+y^2}$ and apply a general formula to find the eccentricity in terms of the conic coefficients. Although I found the correct answer for $\varepsilon$, the calculations were tedious and I have the intuition that there is a simpler way to deduce it. In the case of the semi-major axis $a$, I haven't found a way to compute it. So in conclusion, I would aprecciate that someone helps me in understanding how to deduce this two quantities (and maybe give me some intrinsics of the geometry behind).
Thank you very much!
PD: This problem comes from an approach to solve the 2-body problem of celestial mechanics if anyone is curious.
Best Answer
Substituting polar coordinates $x = \rho \cos \theta$ and $y = \rho \sin \theta$ yields the equation \begin{equation} A \rho \cos \theta + B \rho \sin \theta + \rho = c^2 \end{equation} The major axis is attained at $\theta_0$ and $\theta_0 + \pi$ for which $\rho'(\theta_0)=0$, since these yield the maximum and minimum radii. Implicit differentiation of the above equation and substitution of $\rho'(\theta_0)=0$ yields \begin{equation} -A\sin\theta_0+B\cos\theta_0=0 \end{equation} hence $\tan\theta_0=B/A$. Substitute into the original equation and solve for $\rho(\theta_0)$ to obtain \begin{equation} \rho(\theta_0)=\frac{c^2}{1-\sqrt{1+2hc^2}} \end{equation} \begin{equation} \rho(\theta_0+\pi)=\frac{c^2}{1+\sqrt{1+2hc^2}} \end{equation} The major axis is therefore \begin{equation} 2a = \rho(\theta_0)+\rho(\theta_0+\pi)=-\frac{1}{h} \end{equation} On the other hand, the distance between the two foci is \begin{equation} 2d = \rho(\theta_0)-\rho(\theta_0+\pi)=-\frac{\sqrt{1+2hc^2}}{h} \end{equation} so the eccentricity is the ratio \begin{equation} \varepsilon=\frac{2d}{2a}=\sqrt{1-2hc^2} \end{equation}