# 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?

Dec 1, 2016

This is an ellipse ${\left(x + 1\right)}^{2} / {\left(\sqrt{2}\right)}^{2} + {\left(y - 2\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

#### Explanation:

Let's rearrange the equation

$5 {x}^{2} - {x}^{2} + 6 x + 2 x + {y}^{2} - 4 y = 0$

$4 {x}^{2} + 8 x + {y}^{2} - 4 y = 0$

Completing the squares

$4 \left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 4 y + 4\right) = 8$

$4 {\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 8$

Dividing by $8$

${\left(x + 1\right)}^{2} / \left(2\right) + {\left(y - 2\right)}^{2} / 8 = 1$

${\left(x + 1\right)}^{2} / {\left(\sqrt{2}\right)}^{2} + {\left(y - 2\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

This is the equation of an ellipse

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $\left(- 1 , 2\right)$

$c = \sqrt{{b}^{2} - {a}^{2}} = \sqrt{6}$

The foci are F$\left(- 1 , 2 + \sqrt{6}\right)$ and F'$\left(- 1 , 2 - \sqrt{6}\right)$

The major axis is $= 2 \sqrt{6}$

The minor axis is $= 2 \sqrt{2}$

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