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#?

1 Answer
Oct 27, 2016

Answer:

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)#