# How do you find an equation of hyperbola with given foci and asymptote: Foci: (0,0) and (0,4) Asymptote: y= +/-[(1/2)]x + 2?

Aug 9, 2017

The equation of the hyperbola is ${\left(y - 2\right)}^{2} - \left({x}^{2} / 4\right) = 1$

#### Explanation:

The foci are $F = \left(0 , 4\right)$ and $F ' = \left(0 , 0\right)$

The center is $C = \left(0 , 2\right)$

The equations of the asymptotes are

$y = \frac{1}{2} x + 2$ and $y = - \frac{1}{2} x + 2$

Therefore,

$y - 2 = \pm \frac{1}{2} x$

Squaring both sides

${\left(y - 2\right)}^{2} - \left({x}^{2} / 4\right) = 0$

Therefore,

The equation of the hyperbola is

${\left(y - 2\right)}^{2} - \left({x}^{2} / 4\right) = 1$

Verification

The general equation of the hyperbola is

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

The foci are $F = \left(k , h + c\right) = \left(0 , 2 + 2\right) = \left(0 , 4\right)$ and

$F ' = \left(k , h - c\right) = \left(0 , 2 - 2\right) = \left(0 , 0\right)$

graph{((y-2)^2-(x^2)/4-1)(y-2-1/2x)(y-2+1/2x)=0 [-6.76, 7.28, -1.425, 5.6]}