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

1 Answer
Jul 30, 2018

Answer:

Please see the explanation below

Explanation:

The equation of a hyperbola with a horizontal transverse axis is

#(x-h)^2/a^2-(y-k)^2/b^2=1#

Our equation is

#(x+2)^2/4-(y-5)^2/25=1#

#a=2#

#b=5#

#h=-2#

#k=5#

#c=sqrt(b^2+c^2)=sqrt(25+4)=sqrt29#

The center is #C=(h,k)=(-2,5)#

The vertices are #A=(h+a,k)=(0,5)#

and

#C'=(h-a,k)=(-4,5)#

The foci are #F=(h+c,k)=(-2+sqrt29, 5)#

and

#F'=(-2-sqrt29, 5)#

The asymptotes are

#(x+2)^2/4-(y-5)^2/25=0#

#(x+2)^2/4=(y-5)^2/25#

#(y-5)/5=+-(x+2)/2#

graph{((x+2)^2/4-(y-5)^2/25-1)((y-5)/5-(x+2)/2)((y-5)/5+(x+2)/2)=0 [-12.25, 7.75, 0.2, 10.2]}