# How do you write an equation for the hyperbola with Foci: (0,0) and (0,4) Asmyptotes: y= +-1/2x + 2?

Jun 3, 2018

Equation of vertical hyperbola is $\frac{5 {\left(y - 2\right)}^{2}}{4} - \frac{5 {x}^{2}}{16} = 1$

#### Explanation:

Focii of hyperbola is at $\left(0 , 0\right) \mathmr{and} \left(0 , 4\right)$

Center of the hyperbola is at $\left(0 , 2\right) \therefore \left(h = 0 , k = 2\right)$

So transverse axis is vertical , and equation of asymptotes

of vertical hyperbola are $y - k = \pm \frac{a}{b} \left(x - h\right)$

Asymptotes:  y= +- 1/2 x+2; m=1/2 :. a/b= 1/2 :. b=2 a

We know c^2= a^2+b^2 ; c=4-2=2

$\therefore {2}^{2} = {a}^{2} + {b}^{2} \mathmr{and} 4 = {a}^{2} + {\left(2 a\right)}^{2} \mathmr{and} 5 {a}^{2} = 4$ or

${a}^{2} = \frac{4}{5} \mathmr{and} a = \frac{2}{\sqrt{5}} \therefore b = 2 a = \frac{4}{\sqrt{5}}$

The standard form of equation of vertical hyperbola is

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

${\left(y - 2\right)}^{2} / \left(\frac{4}{5}\right) - {\left(x - 0\right)}^{2} / \left(\frac{16}{5}\right) = 1$ or

$\frac{5 {\left(y - 2\right)}^{2}}{4} - \frac{5 {x}^{2}}{16} = 1$

graph{5(y-2)^2/4-5 x^2/16=1 [-12.66, 12.65, -6.33, 6.33]}