# 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?

Jan 11, 2017

It is an ellipse.

#### Explanation:

Let us write the equation in the form $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

As $5 {x}^{2} + 6 x - 4 y = {x}^{2} - {y}^{2} - 2 x$ is equivalent to

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

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

Hence, here we have $A = 4$, $B = 0$ and $C = 1$

and therefore ${B}^{2} - 4 A C = 0 - 4 \times 4 \times 1 = - 16 < 0$ and $A \ne C$

Hence it is an ellipse (for details see here).

The equation can be written as $4 \left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 4 y + 4\right) = 8$

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

or ${\left(x + 1\right)}^{2} / 2 + {\left(y - 2\right)}^{2} / 8 = 1$ and comparing it with general form of equation of an ellipse ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$, where center is $\left(h , k\right)$ and axes of symmetry are $x = h$ and $y = k$. The two axes are $2 a$ and $2 b$.

Here, we have center at $\left(- 1 , 2\right)$ and axes of symmetry are $x = - 1$ and $y = 2$. It has major axis $2 \sqrt{8} = 4 \sqrt{2}$ (vertically aligned as $b > a$) and minor axis is $2 \sqrt{2}$.

graph{5x^2+6x-4y=x^2-y^2-2x [-11.75, 8.25, -2.84, 7.16]}