How do you find the center, vertices, foci and asymptotes of #4x^2-y^2-24x-4y+28=0#?

1 Answer

Center #(h, k)=(3, -2)#
Vertex #(h+a, k)=(4, -2)# and #(h-a, k)=(2, -2)#
Foci #(h+c, k)=(5.23, -2)# and #(h-c, k)=(0.77, -2)#
Asymptotes #y=2x-8# and #y=-2x+4#

Explanation:

From the given equation
#4x^2-y^2-24x-4y+28=0#

rearrange first so that the variables are together

#4x^2-24x-y^2-4y+28=0#

Perform completing the square

#4(x^2-6x)-(y^2+4y)+28=0#

#4(x^2-6x+9-9)-(y^2+4y+4-4)+28=0#

#4(x-3)^2-36-(y+2)^2+4+28=0#

#4(x-3)^2-(y+2)^2-4=0#

#(x-3)^2/1-(y+2)^2/4=1#

From the form of hyperbola

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

#a=1# and #b=2# and #c=sqrt5#

Center #(h, k)=(3, -2)#

Vertex
#(h+a, k)=(3+1, -2)=(4, -2)#
#(h-a, k)=(3-1, -2)=(2, -2)#

Foci
#(h+c, k)=(3+sqrt5, -2)=(5.23, -2)#
#(h-c, k)=(3-sqrt5, -2)=(0.77, -2)#

Asymptotes

#y=+-b/a(x-h)+k#

#y=+2/1(x-3)+(-2)#
#y=2x-6-2#
#y=2x-8#
and

#y=-2/1(x-3)+(-2)#

#y=-2x+6-2#

#y=-2x+4#

kindly see the graph

hyperbola-colored red
asymptotes-colored blue
center-colored violet
vertices-colored orange
foci-colored black

enter image source here

God bless...I hope the explanation is useful.