How do you identity if the equation #5x^2+6x-4y=x^2-y^2-2x# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Dec 1, 2016

Answer:

This is an ellipse #(x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1#

Explanation:

Let's rearrange the equation

#5x^2-x^2+6x+2x+y^2-4y=0#

#4x^2+8x+y^2-4y=0#

Completing the squares

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

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

Dividing by #8#

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

#(x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1#

This is the equation of an ellipse

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

The center is #(-1,2)#

#c=sqrt(b^2-a^2)=sqrt6#

The foci are F#(-1,2+sqrt6)# and F'#(-1,2-sqrt6)#

The major axis is #=2sqrt6#

The minor axis is #=2sqrt2#

graph{4x^2+8x+y^2-4y=0 [-6.99, 7.06, -1.27, 5.754]}