Determine equation in standard form of a hyperbola with foci at $(3,7)$ and $(3,-1)$ with eccentricity $e=2$.
Working:
The equation must be in form $\frac{(y-q)^2}{a^2}-\frac{(x-s)^2}{b^2}=1$, where $a^2+b^2=c^2$.
The foci lie on the line $x=3$ which is the major axis. The center is the midpoint of the foci at $(3,3)$. So $q=3$ and $s=3$.
Normally for the major axis being parallel to the y-axis, the foci would be at $(0,c)$ and $(0,-c)$. So in this case they are at $(3,c+3)$ and $(3,-c+3)$. This means $c=4$.
Since $e=\frac{c}{a}$, we must have $a=2$ and $b=2 \sqrt{3}$. This gives $\frac{(y-3)^2}{4}-\frac{(x-3)^2}{12}=1$.
However, I think this is wrong as the distance from a point to the two foci must always add to $2a$ which isn't the case. I know my center is correct though.
Best Answer
Your answer $\frac{(y-3)^2}{4}-\frac{(x-3)^2}{12}=1$ is correct.
No, for your hyperbola (not ellipse), we have $|PF_1-PF_2|=4$.