Question

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?

Answer 1

The equation of the hyperbola is `(y-2)^2-(x^2/4)=1`

Explanation:

The foci are `F=(0,4)` and `F'=(0,0)`

The center is `C=(0,2)`

The equations of the asymptotes are

`y=1/2x+2` and `y=-1/2x+2`

Therefore,

`y-2=+-1/2x`

Squaring both sides

`(y-2)^2-(x^2/4)=0`

Therefore,

The equation of the hyperbola is

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

Verification

The general equation of the hyperbola is

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

The foci are `F=(k,h+c)=(0,2+2)=(0,4)` and

`F'=(k,h-c)=(0,2-2)=(0,0)`

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]}