# Find the equation of hyperbola whose foci are $F_1 = (3, 4)$ and $F_2 = (-1,-2)$ and $a=1$?

Find the equation of hyperbola whose foci are $F_1 = (3, 4)$ and $F_2 = (-1,-2)$ and $a=1$?

I need some help with this exercise. I know that this hyperbola is not centered at the origin, but I don't know its orientation and consequently the form of its equation. Does the distance between the foci still $2c$ in this case?

• Hint: What the midpoint of the two foci? – lab bhattacharjee Dec 9 '17 at 16:41
• – lab bhattacharjee Dec 9 '17 at 16:45
• @labbhattacharjee It's not that simple. The hyperbola is $45^\circ$ rotated. This will need a bit more work hence the potential cause why the OP is stuck. – Karn Watcharasupat Dec 9 '17 at 16:48

Apply the definition, directly

$PF_1-PF_2=2a$

$$\sqrt{(x-3)^2+(y-4)^2}-\sqrt{(x+1)^2+(y+2)^2}=2$$ $$\sqrt{x^2-6 x+y^2-8 y+25}=\sqrt{x^2+2 x+y^2+4 y+5}+2$$

and square both sides $$x^2-6 x+y^2-8 y+25=x^2+2 x+y^2+4 y+9+4 \sqrt{(x+1)^2+(y+2)^2}$$ Rearrange and divide both sides by $4$ $$4 - 2 x - 3 y=\sqrt{x^2+2 x+y^2+4 y+5}$$ and square again $$\color{red}{3 x^2+8 y^2+12 x y-18 x-28 y+11=0}$$ is the equation of the wanted hyperbola.

Hope this can be useful

$$...$$

• Thank u so much, you saved my semester,hahah! – Jessy Dec 9 '17 at 20:39

We can easily find the center of the hyperbola to be $(1,1)$ which is the midpoint of the foci and the slope of the major axis is $\frac{3}{2}$ since the major axis passes through both foci.

We still have $c=\sqrt{13}$ as per usual since rotation does not affect the defining properties of the hyperbola or a conic for that matter. Hence, $b=\sqrt{13-1}=\sqrt{12}$.

So we know that the hyperbola is the curve $C_0$ rotated by $\arctan\left(\frac{3}{2}\right)$ counterclockwise and translated by $(1,1)$ about the origin, where $C_0$: $$x_0^2-\frac{y_0^2}{12}=1$$

I will leave the manipulations out. Eventually the curve should have the equation: $$3x^2-18x+12xy-28y+ 8y^2+11=0$$