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

1 Answer
Nov 4, 2016

Answer:

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]}