# How do you write an equation of an ellipse in standard form passing through the center at the origin that passes through the points (1, (-10√2)/3 ) and (-2, (5√5)/3 )?

Jan 28, 2017

The standard form of the equation of an ellipse is:
${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ [1]}$
where $\left(h , k\right)$ is the center.

#### Explanation:

We are given that the center is the origin, $\left(0 , 0\right)$, therefore, we can substitute 0 for h and 0 for k into equation [1] to give us equation [2]:

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

Use the two given points and equation [2] to write two equations:

${\left(1 - 0\right)}^{2} / {a}^{2} + {\left(\frac{- 10 \sqrt{2}}{3} - 0\right)}^{2} / {b}^{2} = 1 \text{ [3]}$

${\left(- 2 - 0\right)}^{2} / {a}^{2} + {\left(\frac{5 \sqrt{5}}{3} - 0\right)}^{2} / {b}^{2} = 1 \text{ [4]}$

Let $u = \frac{1}{a} ^ 2$, let $v = \frac{1}{b} ^ 2$ and square the numerators:

$u + \left(\frac{200}{9}\right) v = 1 \text{ [5]}$

$4 u + \left(\frac{125}{9}\right) v = 1 \text{ [6]}$

Convert to an augmented matrix :

[(1, 200/9,|,1), (4, 125/9,|,1) ]

Perform Elementary row operations :

${R}_{2} - 3 {R}_{1} \to {R}_{2}$

[(1, 200/9,|,1), (0, -675/9,|,-3) ]

$- 9 {R}_{2} / 675$

[(1, 200/9,|,1), (0, 1,|,1/25) ]

${R}_{1} - \frac{200}{9} {R}_{2} \to {R}_{1}$

[(1, 0,|,1/9), (0, 1,|,1/25) ]

$u = \frac{1}{9} \mathmr{and} v = \frac{1}{25}$

$\frac{1}{a} ^ 2 = \frac{1}{9} \mathmr{and} \frac{1}{b} ^ 2 = \frac{1}{25}$

$a = 3 \mathmr{and} b = 5$

Substitute this into equation [2]:

${\left(x - 0\right)}^{2} / {3}^{2} + {\left(y - 0\right)}^{2} / {5}^{2} = 1 \text{ [7]}$

Equation [7] is the answer.