# What conic section does the equation x^2 + 4y^2 - 4x + 8y - 60 = 0  represent?

Sep 25, 2014

In this problem we are going to rely on the completing the square technique to massage this equation into an equation that is more recognizable.

${x}^{2} - 4 x + 4 {y}^{2} + 8 y = 60$

Let's work with the $x$ term

${\left(- \frac{4}{2}\right)}^{2} = {\left(- 2\right)}^{2} = 4$, We need to add 4 to both sides of the equation

${x}^{2} - 4 x + 4 + 4 {y}^{2} + 8 y = 60 + 4$

${x}^{2} - 4 x + 4 \implies {\left(x - 2\right)}^{2} \implies$Perfect square trinomial

Re-write equation:

${\left(x - 2\right)}^{2} + 4 {y}^{2} + 8 y = 60 + 4$

Let's factor out a 4 from the ${y}^{2}$ & $y$ terms

${\left(x - 2\right)}^{2} + 4 \left({y}^{2} + 2 y\right) = 60 + 4$

Let's work with the $y$ term

${\left(\frac{2}{2}\right)}^{2} = {\left(1\right)}^{2} = 1$, We need to add 1 to both sides of the equation

But remember that we factored out a 4 from the left side of the equation. So on the right side we are actually going to add 4 because $4 \cdot 1 = 4.$

${\left(x - 2\right)}^{2} + 4 \left({y}^{2} + 2 y + 1\right) = 60 + 4 + 4$

${y}^{2} + 2 y + 1 \implies {\left(y + 1\right)}^{2} \implies$Perfect square trinomial

Re-write equation:

${\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 60 + 4 + 4$

${\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 68$

$\frac{{\left(x - 2\right)}^{2}}{68} + \frac{4 {\left(y + 1\right)}^{2}}{68} = \frac{68}{68}$

$\frac{{\left(x - 2\right)}^{2}}{68} + \frac{{\left(y + 1\right)}^{2}}{17} = 1$

This is an ellipse when a center (2,-1).

The $x$-axis is the major axis.

The $y$-axis is the minor axis.