# How do you find the center, vertices, foci and eccentricity of 9x^2+4y^2-36x+8y+31=0?

Aug 13, 2016

Centre: $\left(2 , - 1\right)$

Vertices: $\left(2 , \frac{1}{2}\right) \mathmr{and} \left(2 , - \frac{5}{2}\right)$

Co-Vertices: $\left(1 , - 1\right) \mathmr{and} \left(3 , - 1\right)$

Foci: $\left(2 , \frac{- 2 + \sqrt{5}}{2}\right) \mathmr{and} \left(2 , \frac{- 2 - \sqrt{5}}{2}\right)$

Eccentricity: $\frac{\sqrt{5}}{3}$

#### Explanation:

The technique we want to use is called completing the square. We shall use it on the $x$ terms first and then the $y$.

Rearrange to

$9 {x}^{2} + 4 {y}^{2} - 36 x + 8 y = - 31$

Focussing on $x$, divide through by the ${x}^{2}$ coefficient and add the square of half the coefficient of the ${x}^{1}$ term to both sides:

${x}^{2} + \frac{4}{9} {y}^{2} - 4 x + \frac{8}{9} y + {\left(- 2\right)}^{2} = - \frac{31}{9} + {\left(- 2\right)}^{2}$

${\left(x - 2\right)}^{2} + \frac{4}{9} {y}^{2} + \frac{8}{9} y = \frac{5}{9}$

Divide through by ${y}^{2}$ coefficient and add square of half the coefficient of the ${y}^{1}$ term to both sides:

$\frac{9}{4} {\left(x - 2\right)}^{2} + {y}^{2} + 2 y + {\left(1\right)}^{2} = \frac{5}{4} + {\left(1\right)}^{2}$

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

Divide by $\frac{9}{4}$ to simplify:

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

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

General equation is

${\left(x - a\right)}^{2} / {h}^{2} + {\left(y - b\right)}^{2} / {k}^{2} = 1$

where $\left(a , b\right)$ is the centre and $h , k$ are the semi-minor/major axis.

Reading off the centre gives $\left(2 , - 1\right)$.

In this case, the $y$ direction has a bigger value than the $x$, so the ellipse will be stretched in the $y$ direction. ${k}^{2} > {h}^{2}$

The vertices are obtained by moving up the major axis from the centre. Ie $\pm \sqrt{k}$ added to the y coordinate of the centre.

This gives $\left(2 , \frac{1}{2}\right) \mathmr{and} \left(2 , - \frac{5}{2}\right)$.

The co-vertices lie on the minor axis. We add $\pm \sqrt{h}$ to the centre's x coordinate to find these.

$\left(1 , - 1\right) \mathmr{and} \left(3 , - 1\right)$

Now, to find the foci:

${c}^{2} = {k}^{2} - {h}^{2}$

${c}^{2} = \frac{9}{4} - 1$

${c}^{2} = \frac{5}{4} \implies c = \pm \frac{\sqrt{5}}{2}$

Foci will be situated along the line $x = 2$ at $\pm \frac{\sqrt{5}}{2}$ from $y = - 1$.

$\therefore$ foci at $\left(2 , \frac{- 2 + \sqrt{5}}{2}\right) \mathmr{and} \left(2 , \frac{- 2 - \sqrt{5}}{2}\right)$

Finally the eccentricity is found using

$e = \sqrt{1 - {h}^{2} / {k}^{2}}$

$e = \sqrt{1 - \frac{1}{\frac{9}{4}}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$