Finding the equation of hyperbola that passes through $(2,0)$ and has foci points $(2,3)$ and $(1,0)$

Question

If the hyperbola is given by $$(x-x_0)^2/a^2-(y-y_0)^2/b^2=1$$ and I know that if $F(c,0)$ is a foci point then $c^2=a^2+b^2$ I can say that in this particular problem $a^2+b^2=1$.Then I plug in the given point and I get $$(2-x_0)^2/a^2-(y_0)^2/1-a^2=1$$ Now I have one equation with two unknowns. Could you please explain how to solve this problem. Thank you.

Answer 1

The center of the hyperbola is the midpoint of the segment joining $F_1(2,3),F_2(1,0)$ that is $M\left(\frac32,\frac32\right)\equiv(x_0,y_0)$.

Thus $c^2=MF_1^2$

This is a rotated hyperbola.

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Edit

$(x_0,y_0)=\left(\frac32,\frac32\right)$

$c^2=MF_1^2=5/2$

$$\frac{\left(2-\frac{3}{2}\right)^2}{a^2}-\frac{\left(0-\frac{3}{2}\right)^2}{\frac{5}{2}-a^2}=1$$ solve for $a$

Answer 2

Using A hyperbola as a constant difference of distances if $P(h,k)$ is an arbitrary point of the required hyperbola,

$$2a=|\sqrt{(h-2)^2+(k-3)^2}-\sqrt{(h-1)^2+(k-0)^2}|, a>0$$

As the hyperbola passes through $(2,0);$

$$2a=|\sqrt{(2-2)^2+(0-3)^2}-\sqrt{(2-1)^2+(0-0)^2}|=?$$