# How do I graph the ellipse with the equation x^2+4y^2-4x+8y-60=0?

Jun 6, 2016

An ellipse centered at (2, –1) with horizontal radius $\sqrt{68}$ units and vertical radius $\sqrt{17}$ units.

#### Explanation:

Firstly, complete the square.

${x}^{2} - 4 x + 4 {y}^{2} + 8 y = 60$
${x}^{2} - 2 \left(2\right) x + {2}^{2} + 4 \left({y}^{2} + 2 \left(1\right) y + {1}^{2}\right) = 60 + {2}^{2} + 4$
${\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 68$
${\left(x - 2\right)}^{2} / 68 + {\left(y + 1\right)}^{2} / 17 = 1$
${\left(x - 2\right)}^{2} / {\left(\sqrt{68}\right)}^{2} + {\left(y + 1\right)}^{2} / {\left(\sqrt{17}\right)}^{2} = 1$

Thus the graph is an ellipse centered at (2, –1) with horizontal radius $\sqrt{68}$ units and vertical radius $\sqrt{17}$ units.