# What does the equation (x-1)^2/4-(y+2)^2/9=1 tell me about its hyperbola?

Jul 23, 2018

#### Explanation:

The general equation of a hyperbola is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

Here,

The equation is

${\left(x - 1\right)}^{2} / {2}^{2} - {\left(y + 2\right)}^{2} / {3}^{2} = 1$

$a = 2$

$b = 3$

$c = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{4 + 9} = \sqrt{13}$

The center is $C = \left(h , k\right) = \left(1 , - 2\right)$

The vertices are

$A = \left(h + a , k\right) = \left(3 , - 2\right)$

and

$A ' = \left(h - a , k\right) = \left(- 1 , - 2\right)$

The foci are

$F = \left(h + c , k\right) = \left(1 + \sqrt{13} , - 2\right)$

and

$F ' = \left(h - c , k\right) = \left(1 - \sqrt{13} , - 2\right)$

The eccentricity is

$e = \frac{c}{a} = \frac{\sqrt{13}}{2}$

graph{((x-1)^2/4-(y+2)^2/9-1)=0 [-14.24, 14.25, -7.12, 7.12]}

#### Explanation:

The given equation of hyperbola

$\setminus \frac{{\left(x - 1\right)}^{2}}{4} - \setminus \frac{{\left(y + 2\right)}^{2}}{9} = 1$

$\setminus \frac{{\left(x - 1\right)}^{2}}{{2}^{2}} - \setminus \frac{{\left(y + 2\right)}^{2}}{{3}^{2}} = 1$

The above equation is in standard form of hyperbola:

${\left(x - {x}_{1}\right)}^{2} / {a}^{2} - {\left(y - {y}_{1}\right)}^{2} / {b}^{2} = 1$

Which has

Eccentricity: $e = \setminus \sqrt{1 + {b}^{2} / {a}^{2}} = \setminus \sqrt{1 + \frac{9}{4}} = \setminus \frac{\sqrt{13}}{2}$

Center: $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(1 , - 2\right)$

Vertices: $\left({x}_{1} \setminus \pm a , {y}_{1}\right) \setminus \equiv \left(1 \setminus \pm 2 , - 2\right)$ &

$\left({x}_{1} , {y}_{1} \setminus \pm b\right) \setminus \equiv \left(1 , - 2 \setminus \pm 3\right)$

Asymptotes: $y - {y}_{1} = \setminus \pm \frac{b}{a} \left(x - {x}_{1}\right)$

$y + 2 = \setminus \pm \frac{3}{2} \left(x - 1\right)$