How do you identity if the equation 5x^2+6x-4y=x^2-y^2-2x5x2+6x4y=x2y22x is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Dec 1, 2016

This is an ellipse (x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1(x+1)2(2)2+(y2)2(8)2=1

Explanation:

Let's rearrange the equation

5x^2-x^2+6x+2x+y^2-4y=05x2x2+6x+2x+y24y=0

4x^2+8x+y^2-4y=04x2+8x+y24y=0

Completing the squares

4(x^2+2x+1)+(y^2-4y+4)=84(x2+2x+1)+(y24y+4)=8

4(x+1)^2+(y-2)^2=84(x+1)2+(y2)2=8

Dividing by 88

(x+1)^2/(2)+(y-2)^2/8=1(x+1)22+(y2)28=1

(x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1(x+1)2(2)2+(y2)2(8)2=1

This is the equation of an ellipse

(x-h)^2/a^2+(y-k)^2/b^2=1(xh)2a2+(yk)2b2=1

The center is (-1,2)(1,2)

c=sqrt(b^2-a^2)=sqrt6c=b2a2=6

The foci are F(-1,2+sqrt6)(1,2+6) and F'(-1,2-sqrt6)(1,26)

The major axis is =2sqrt6=26

The minor axis is =2sqrt2=22

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