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

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Answer 1 · 2016-10-30T02:41:51

The standard form is:

$(x - 0)^2/(sqrt(12))^2 + (y - 0)^2/(sqrt(8))^2 = 1$

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

$(x - h)^2/a^2 + (y - k)^2/b^2 = 1$

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

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

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

$(sqrt(6)- 0)^2/a^2 + (2- 0)^2/b^2 = 1$ [1]
$(-3- 0)^2/a^2 + (sqrt(2)- 0)^2/b^2 = 1$ [2]

Do the multiplication implied by the squares:

$6/a^2 + 4/b^2 = 1$ [3]
$9/a^2 + 2/b^2 = 1$ [4]

Let $u = 1/a^2$ and let $v = 1/b^2$

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

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

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

$-12u = -1$

$u = 1/12$

Substitute $1/12$ for u in equation [6]

$9/12 + 2v = 1$

$2v = 3/12$

$v = 3/24 = 1/8$

$a = sqrt(12) and b = sqrt(8)$

The standard form is:

$(x - 0)^2/(sqrt(12))^2 + (y - 0)^2/(sqrt(8))^2 = 1$

check:

$(sqrt(6) - 0)^2/(sqrt(12))^2 + (2 - 0)^2/(sqrt(8))^2 = 1$
$(-3 - 0)^2/(sqrt(12))^2 + (sqrt(2) - 0)^2/(sqrt(8))^2 = 1$

$6/12 + 4/8 = 1$
$9/12 + 2/8 = 1$

$1 = 1$
$1 = 1$

This checks.