Problem:
Find the equation for the ellipse that has foci
$$F_1 = (0, 0)$$ $$F_2 = (1,1)$$
and eccentricity $$\varepsilon = \frac12.$$
Hint: Use a rotation that moves the foci to the x-axis.
My attempt:
I by rotating the ellipse by $-\pi/4$, the foci are both on the x-axis, with $F_2 = (\sqrt2, 0)$.
The center bisects the line segment between the foci, and is as such $(h,k) = (\sqrt2/2, 0)$. This also gives $c=\sqrt2/2$ as this is the distance between the center and each focus.
The semi-major axis is horizontal, so the $a^2$ goes beneath the $x$-term.
We know that $c = \frac\varepsilon a$, so $a = \frac\varepsilon c = \frac{1}{\sqrt2}$ and $a^2 = \frac12$.
$b^2 = a^2- c^2 = (\frac{1}{\sqrt2})^2 – (\frac{\sqrt2}{2})^2 = 0$
How do I go from here? With $b^2 = 0$, how can I place it in a denominator? Did I make a mistake somewhere?
Best Answer
The co-ordinates of the focii are $(h\pm ae)$, so your $c$ will be $ae$ rather than $\frac{e}{a}$ and so your $a$ will be $\sqrt2$ and your $b$ will be $\sqrt\frac{3}{2}$. So the equation we have in the rotated system is:$$\frac{(x-\frac{1}{\sqrt2})^2}{2}+\frac{2y^2}{3}=1$$ Now all that remains is to somehow rotate this back into the original system.