How do you find all the critical points to graph #(x+3)^2/25-(y+5)^2/4=1# including vertices, foci and asymptotes?

1 Answer
Oct 29, 2016

Answer:

The vertices are #(2,-5)# and #(-8,-5)#
The equations of the asymptotes are #y=-5+2/5(x+3)# and #y=-5-2/5(x+3)#
The foci are #(-3+sqrt29,-5)# and #(-3-sqrt29,-5)#

Explanation:

Our equation is #(x+3)^2/25-(y+5)^2/4=1#
This is the equation of a left to right hyperbola
The typical equation of a hyperbola is #(x-h)^2/a^2-(y-k)^2/b^2=1#
The center is #(h,k)=(-3,-5)#
The vertices are #(h+a,k)=(-3+5,-5)=(2,-5)# and #(h-a,k)=(-3-5,-5)=(-8,-5)#
The slope of the asymptotes are #b/a=2/5# and #-b/a=-2/5#

The equation of the asymptotes are #y=k+b/a(x-h)# and #y-b/a(x-h)#

So the equations are #y=-5+2/5(x+3)# and #y=-5-2/5(x+3)#

In order to calculate the foci, we need #c=+-sqrt(a^2+b^2)=+-sqrt(25+4)=+-sqrt29#
And the foci are #(-3+sqrt29,-5)# and #(-3-sqrt29,-5)#

graph{(x+3)^2/25-(y+5)^2/4=1 [-20.91, 18.56, -13.78, 5.97]}