# How do you find the center, vertices, foci and eccentricity of 4x² + 9y² - 8x - 36y + 4 = 0?

Feb 5, 2016

The centre is at $\left(1 , 2\right)$
The vertices are at $\left(\frac{1}{2} , 2\right) , \left(\frac{3}{2} , 2\right) , \left(1 , 2 \cdot \frac{1}{3}\right)$ and $\left(1 , 1 \cdot \frac{2}{3}\right)$
The foci are at $\left(1 - \frac{\sqrt{5}}{6} , 2\right)$ and $\left(1 + \frac{\sqrt{5}}{6} , 2\right)$

The eccentricity is $\left(\frac{\sqrt{5}}{6}\right)$

#### Explanation:

Rearrange the elements to get the standard form of an ellipse
${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ where $\left(h , k\right)$ is the centre and $a$ and $b$ are the semi-minor and semi-major axes respectively.

So in this example
$4 {x}^{2} - 8 x + 9 {y}^{2} - 36 y + 4 = 0$

Using the method of completing squares

$4 \left({x}^{2} - 2 x\right) + 9 \left({y}^{2} - 4 y\right) + 4 = 0$
$4 {\left(x - 1\right)}^{2} - 1 + 9 {\left(y - 2\right)}^{2} - 4 + 4 = 0$
$4 {\left(x - 1\right)}^{2} + 9 {\left(y - 2\right)}^{2} = 1$
${\left(x - 1\right)}^{2} / \left(\frac{1}{4}\right) + {\left(y - 2\right)}^{2} / \left(\frac{1}{9}\right) = 1$

Therefore the centre is at $\left(1 , 2\right)$
The semi-major axis has length $\frac{1}{2}$ and the semi-minor axis has length $\frac{1}{3}$ so the vertices are at $\left(\frac{1}{2} , 2\right) , \left(\frac{3}{2} , 2\right) , \left(1 , 2 \cdot \frac{1}{3}\right)$ and $\left(1 , 1 \cdot \frac{2}{3}\right)$

If $c$ is the distance between the foci,
$\frac{1}{4} - {c}^{2} = \frac{1}{9}$
$\therefore {c}^{2} = \frac{5}{36}$ and $c = \frac{\sqrt{5}}{6}$

The foci are at $\left(1 - \frac{\sqrt{5}}{6} , 2\right)$ and $\left(1 + \frac{\sqrt{5}}{6} , 2\right)$

The eccentricity is $\frac{c}{a} = \left(\frac{\sqrt{5}}{6}\right)$