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

Question

1582 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T15:47:17

This is an ellipse $\frac{{(x + 1)}^{2}}{{(\sqrt{2})}^{2}} + \frac{{(y - 2)}^{2}}{{(\sqrt{8})}^{2}} = 1$ $(x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1$

Let's rearrange the equation

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

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

Completing the squares

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

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

Dividing by $8$ $8$

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

$\frac{{(x + 1)}^{2}}{{(\sqrt{2})}^{2}} + \frac{{(y - 2)}^{2}}{{(\sqrt{8})}^{2}} = 1$ $(x+1)^2/(sqrt2)^2+(y-2)^2/(sqrt8)^2=1$

This is the equation of an ellipse

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

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

$c = \sqrt{b^{2} - a^{2}} = \sqrt{6}$ $c=sqrt(b^2-a^2)=sqrt6$

The foci are F $(- 1, 2 + \sqrt{6})$ $(-1,2+sqrt6)$ and F' $(- 1, 2 - \sqrt{6})$ $(-1,2-sqrt6)$

The major axis is $= 2 \sqrt{6}$ $=2sqrt6$

The minor axis is $= 2 \sqrt{2}$ $=2sqrt2$

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

How do you identity if the equation 5x2+6x−4y=x2−y2−2x5x^2+6x-4y=x^2-y^2-2x is a parabola, circle, ellipse, or hyperbola and how do you graph it?