Question

How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola `4x^2-25y^2-8x-96=0`?

Answer 1

The vertices are `(6,0)` and `(-4,0)`
The equations of the asymptotes are `y=2/5(x-1)` and `y=-2/5(x-1)`
The foci are `(1+sqrt29,0)` and `(1-sqrt29,0)`

Explanation:

The equation is `4x^2-8x-25y^2=96`

Completing the square,give

`4(x^2-2x)-25y^2=96`

`4(x^2-2x+1)-25y^2=96+4`

`4(x-1)^2-25y^2=100`
Dividing by 100

`(x-1)^2/25-y^2/4=1`

`(x-1)^2/5^2-y^2/2^2=1`
This is the standard equation of a left right hyperbola
`(x-h)^2/a^2-(y-k)^2/b^2=1`

so, the center is `(h,k)=(1,0)`
the vertices ae `(h+a,k)=(1+5,0)=(6,0)`
and `(h-a,k)=(1-5,0)=(-4,0)`
The slopes of the asymptotes are `+-b/a=+-2/5`
And the equations of the asymptotes are `y=+-b/a(x-h)`
`=>` `y=+-2/5(x-1)`
To determine the foci, we need `c=+-sqrt(a^2+b^2)=+-sqrt(25+4)=+-sqrt29`
So the foci are `(h+c,k)=(1+sqrt29,0)`
and `(h-c,k)=(1-sqrt29,0)`