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.
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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.
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 $a$
Raffaele
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Can I say that $a^2+b^2=1/2$? for a rotated hyperbola and then plug in the given point and solve for $x_0$ and $y_0$ ? – Britanica Dec 24 '20 at 14:53
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@Britanica I added some detail to my answer – Raffaele Dec 24 '20 at 15:07
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I get four crazy possible values foe $a$ e.g. $$sqrt(10-3sqrt10)/2$$. Is that fine? – Britanica Dec 24 '20 at 15:21
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1Result is $$\frac{4 \left(x-\frac{3}{2}\right)^2}{10-3 \sqrt{10}}-\frac{2}{3} \sqrt{\frac{2}{5}} \left(y-\frac{3}{2}\right)^2=1$$ – Raffaele Dec 24 '20 at 15:24
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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 $(2,0);$
$$2a=|\sqrt{(2-2)^2+(0-3)^2}-\sqrt{(2-1)^2+(0-0)^2}|=?$$
lab bhattacharjee
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@Britanica, How $b=0?$ Have you reached the equation of the hyperbola? – lab bhattacharjee Dec 24 '20 at 15:00
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1@Britanica, $$2ae=\sqrt{(2-1)^2+(3-0)^2}=\sqrt{10}$$ and $$2a=1$$ $$\implies e=?$$ Finally $$b^2=a^2(e^2-1)=\dfrac{10-1}4$$ – lab bhattacharjee Dec 24 '20 at 15:06