If I know the coordinates of the foci F1, F2 and the coordinate of a vertex P1 that lies on the hyperbola (both expressed in 2D cartesian coordinates).
How would I determine the equation of the hyperbola.
Note that the line that passes through F1, and F2 may not always be parallel with the X/Y axis.
Best Answer
For example take $F_1=(0,0),F_2=(3,3),P_1=(1,1):$
$\sqrt{(x-0)^2+(y-0)^2}-\sqrt{(x-3)^2+(y-3)^2}=\pm 2a$
$\sqrt2-\sqrt8=\pm 2a, a^2=\frac12$
$x^2+y^2=(\pm 2a+ \sqrt{(x-3)^2+(y-3)^2 })^2$
$x^2+y^2=4a^2\pm 4a \sqrt{(x-3)^2+(y-3)^2}+(x-3)^2+(y-3)^2$
$\mp 4a \sqrt{(x-3)^2+(y-3)^2 } =4a^2+(x-3)^2+(y-3)^2-x^2-y^2$
$16a^2((x-3)^2+(y-3)^2) =(4a^2-6x+9-6y+9)^2$
$16a^2((x-3)^2+(y-3)^2) -(4a^2-6x-6y+18)^2=0$
$8((x-3)^2+(y-3)^2)-(20-6x-6y)^2=0$
$-28 x^2 - 72 x y - 28 y^2 + 192 x + 192 y - 256=0.$
PS: In a coordinate system keeping the same distances and center $(\frac32,\frac32)$ as the origin and the $x'$-axis along $y=x$ the equation is $\frac{x'^2}{a^2}-\frac{y'^2}{b^2}=1,$ where $a$ is the distance from the vertices to the center, $c$ the distance from the foci to the center; $a^2=\frac12, c^2=\frac92$ and $b^2=c^2-a^2=4.$