Question

How do you find all the critical points to graph `9x^2 - 4y^2 - 90x - 32y + 125 = 0` including vertices, foci and asympotes?

Answer 1

The vertices are `(7,-4)` and `(3,-4)`
The foci are `(5+sqrt13,-4)` and `(5-sqrt13,-4)`
The equations of the asymptotes are `y=-4+3/2(x-5)` and `y=-4-3/2(x-5)`

Explanation:

This is the equation of a hyperbola
Rearranging the terms and completing the squares
`9(x^2-10x)-4(y^2+8y)=-125`
`9(x^2-10x+25)-4(y^2+8y+16)=-125+225-64=36`
`9(x-5)^2-4(y+4)^2=36`
dividing by `36`
`(x-5)^2/4-(y+4)^2/9=1`
The center of this left right hyperbola is `=(5,-4)`
`a=2` and `b=3`
The vertices are `(7,-4)` and `(3,-4)`
The slope of the asymptotes are `+-3/2`
The equations of the asymptotes are `y=-4+3/2(x-5)` and `y=-4-3/2(x-5)`
To calculate the foci, we need `c=sqrt(4+9)=sqrt13`4
The foci are `(5+sqrt13,-4)` and `(5-sqrt13,-4)`
graph{(x-5)^2/4-(y+4)^2/9=1 [-13.86, 14.61, -9.24, 5]}