Equation is 4x²+8y²+8x-4y=8. what is the vertices,co-vertices,foci?

1 Answer
Jul 12, 2018

Please see the explanation below

Explanation:

Let's rearrange the equation

#4x^2+8y^2+8x-4y=8#

#4x^2+8x+8y^2-4y=8#

Dividing by #4#

#x^2+2x+2y^2-y=2#

Completing the square

#x^2+2x+1+2(y^2-1/2y+1/16)=2+1+1/8#

#(x+1)^2+2(y-1/4)^2=25/8#

Dividing by #25/8#

#(x+1)^2/(25/8)+2(y-1/2)^2/(25/8)=1#

#(x+1)^2/(5/sqrt8)^2+(y-1/4)^2/(5/4)^2=1#

The general equation of an ellipse is

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

This is an equation of an ellipse, center #C=(h,k)=(-1,1/4)#

The vertices are #A=(h+a,k)=(-1+5/sqrt8,1/4)=(0.768,0.25)#

#A'=(h-a,k)=((-1)-5/sqrt8, 1/2)=(-2.768,0.25)#

#B=(h, k+b)=(-1, 1/4+5/4)=(-1, 3/2)#

#B'=(h, k-b)=(-1, 1/4-5/4)=(-1, -1)#

Calculate #c#

#c^2=a^2-b^2=25/8-25/16=25/16#

The foci are

#F=(h+c,k)=(-1+5/4, 1/4)=(1/4,1/4)#

And

#F'=(h-c, k), (-1-5/4,1/4)=(-9/4,1/4)#

graph{x^2+2y^2+2x-y-2=0 [-5.308, 2.487, -1.806, 2.09]}