# How do you identity if the equation 7x^2-28x+4y^2+8y=-4 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Dec 28, 2016

It's an ellipse with $x$-semiaxis $2$ and $y$-semiaxis $\sqrt{7}$ centred on the point $\left(2 , - 1\right)$

#### Explanation:

It is probably an ellipse because the coefficients of ${x}^{2}$ and ${y}^{2}$ have the same sign but different values, and there is no term in $x y$.

So complete the square for the $x$ and $y$ separately:
$7 {x}^{2} - 28 x + 4 {y}^{2} + 8 y = - 4$
$7 {\left(x - 2\right)}^{2} - 28 + 4 {\left(y + 1\right)}^{2} - 4 = - 4$
$7 {\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 28$
Then reduce to standard form:
${\left(\frac{x - 2}{2}\right)}^{2} + {\left(\frac{y + 1}{\sqrt{7}}\right)}^{2} = 1$
which is of the standard form:
${\left(\frac{x}{a}\right)}^{2} + {\left(\frac{y}{b}\right)}^{2} = 1$
so you can read off the semi-axes $a$ and $b$ and note that the centre is at $\left(2 , - 1\right)$.