How do you find an equation for the ellipse with vertices at (-6,4) and (10,4); focus at (8,4)?

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Answer 1 · 2016-11-16T01:12:12

Please see the explanation.

The equation of an ellipse:

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

Has vertices at $(h \pm a, k)$ $(h +-a, k)$
Has foci at $(h \pm \sqrt{a^{2} - b^{2}}, k)$ $(h +-sqrt(a^2 -b^2), k)$

Use the vertices to write 3 equations:

$k = 4 [1]$ $k = 4" [1]"$
$h - a = - 6 [2]$ $h - a = -6" [2]"$
$h + a = 10 [3]$ $h + a = 10" [3]"$

Use equations [2] and [3] to solve for h and a:

$2 h = 4$ $2h = 4$
$h = 2$ $h = 2$
$a = 8$ $a = 8$

Use the focus to write another equation:

$8 = h + \sqrt{a^{2} - b^{2}}$ $8 =h + sqrt(a^2 - b^2)$

Substitute values for h and a:

$8 = 2 + \sqrt{8^{2} - b^{2}}$ $8 = 2 + sqrt(8^2 - b^2)$

$6 = \sqrt{64 - b^{2}}$ $6 = sqrt(64 - b^2)$

$36 = 64 - b^{2}$ $36 = 64 - b^2$

$b^{2} = 64 - 36$ $b^2 = 64 - 36$

$b^{2} = 28$ $b^2 = 28$

$b = \sqrt{28}$ $b = sqrt(28)$

Substitute the values into the standard form:

$\frac{{(x - 2)}^{2}}{8^{2}} + \frac{{(y - 4)}^{2}}{{(\sqrt{28})}^{2}} = 1$ $(x - 2)^2/8^2 + (y - 4)^2/(sqrt(28))^2 = 1$