# How do you write an equation of an ellipse in standard form given center at origin and passes through (√6, 2) and (-3, √2)?

##### 1 Answer
Oct 30, 2016

The standard form is:

${\left(x - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 0\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

#### Explanation:

The standard form of an equation of an ellipse with an arbitrary center $\left(h , k\right)$ is:

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

The standard form equation for an ellipse with its center at the origin is:

${\left(x - 0\right)}^{2} / {a}^{2} + {\left(y - 0\right)}^{2} / {b}^{2} = 1$

Write two equations using the above form and the two given points:

${\left(\sqrt{6} - 0\right)}^{2} / {a}^{2} + {\left(2 - 0\right)}^{2} / {b}^{2} = 1$ [1]
${\left(- 3 - 0\right)}^{2} / {a}^{2} + {\left(\sqrt{2} - 0\right)}^{2} / {b}^{2} = 1$ [2]

Do the multiplication implied by the squares:

$\frac{6}{a} ^ 2 + \frac{4}{b} ^ 2 = 1$ [3]
$\frac{9}{a} ^ 2 + \frac{2}{b} ^ 2 = 1$[4]

Let $u = \frac{1}{a} ^ 2$ and let $v = \frac{1}{b} ^ 2$

$6 u + 4 v = 1$ [5]
$9 u + 2 v = 1$ [6]

Multiply equation [6] by -2 and add to equation [5]

$6 u - 18 u + 4 v - 4 v = 1 - 2$

$- 12 u = - 1$

$u = \frac{1}{12}$

Substitute $\frac{1}{12}$ for u in equation [6]

$\frac{9}{12} + 2 v = 1$

$2 v = \frac{3}{12}$

$v = \frac{3}{24} = \frac{1}{8}$

$a = \sqrt{12} \mathmr{and} b = \sqrt{8}$

The standard form is:

${\left(x - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 0\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

check:

${\left(\sqrt{6} - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(2 - 0\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$
${\left(- 3 - 0\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(\sqrt{2} - 0\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

$\frac{6}{12} + \frac{4}{8} = 1$
$\frac{9}{12} + \frac{2}{8} = 1$

$1 = 1$
$1 = 1$

This checks.