Question

How do you write the equation of a hyperbola in standard form given Foci: (3,+-2) and Asymptotes: y = +-2(x-3)?

Answer 1

The equation of the hyperbola is `(y)^2/(16/5)-(x-3)^2/(4/5)=1`

Explanation:

The standard equation of a hyperbola is
`(y-k)^2/b^2-(x-h)^2/a^2=1`
The slopes of the asymptotes are `+-b/a`
so `+-b/a=+-2` `=>` `b=2a`
The foci are `h,k+-c`
So `h=3`
and `k+c=2` and `k-c=-2`
So from the eqautions `k=0` and `c=2`
`c^2=a^2+b^2`
Therefore `4=a^2+4a^2=5a^2` `=>` `a^2=4/5`
And `b^2=4a^2=16/5`

So the equation of the hyperbola is
`(y)^2/(16/5)-(x-3)^2/(4/5)=1`