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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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:

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

Explanation:

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

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

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

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

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

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

Do the multiplication implied by the squares:

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

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

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

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

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

$- 12 u = - 1$ $-12u = -1$

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

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

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

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

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

$a = \sqrt{12} and b = \sqrt{8}$ $a = sqrt(12) and b = sqrt(8)$

The standard form is:

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

check:

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

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

$1 = 1$ $1 = 1$
$1 = 1$ $1 = 1$

This checks.

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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

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