Question

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

Answer 1

Equation of vertical hyperbola is `(5(y-2)^2)/4-(5 x^2)/16=1`

Explanation:

Focii of hyperbola is at `(0,0) and (0,4)`

Center of the hyperbola is at `(0, 2) :. (h=0,k=2)`

So transverse axis is vertical , and equation of asymptotes

of vertical hyperbola are `y-k =+- a/b(x-h)`

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`

`:. 2^2= a^2+b^2 or 4 = a^2+(2 a)^2 or 5 a^2 =4` or

`a^2= 4/5 or a = 2/sqrt 5 :. b= 2a= 4 /sqrt 5`

The standard form of equation of vertical hyperbola is

`(y-k)^2/a^2-(x-h)^2/b^2=1` or

`(y-2)^2/(4/5)-(x-0)^2/(16/5)=1` or

`(5(y-2)^2)/4-(5 x^2)/16=1`

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