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 )?

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Answer 1 · 2017-01-28T18:07:04

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

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

$(x-0)^2/a^2+(y-0)^2/b^2=1" [2]"$

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

$(1-0)^2/a^2 + ((-10sqrt2)/3-0)^2/b^2 = 1" [3]"$

$(-2-0)^2/a^2 + ((5sqrt5)/3-0)^2/b^2 = 1" [4]"$

Let $u = 1/a^2$ , let $v = 1/b^2$ and square the numerators:

$u + (200/9)v = 1" [5]"$

$4u + (125/9)v = 1" [6]"$

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

$R_2-3R_1toR_2$

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

$-9R_2/675$

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

$R_1-200/9R_2toR_1$

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

$u = 1/9 and v = 1/25$

$1/a^2 = 1/9 and 1/b^2 = 1/25$

$a = 3 and b = 5$

Substitute this into equation [2]:

$(x-0)^2/3^2+(y-0)^2/5^2=1" [7]"$

Equation [7] is the answer.