Question

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?

Answer 1

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]}