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

Question

11579 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T01:12:12

Please see the explanation.

The equation of an ellipse:

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

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

Use the vertices to write 3 equations:

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

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

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

Use the focus to write another equation:

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

Substitute values for h and a:

$8 = 2 + sqrt(8^2 - b^2)$

$6 = sqrt(64 - b^2)$

$36 = 64 - b^2$

$b^2 = 64 - 36$

$b^2 = 28$

$b = sqrt(28)$

Substitute the values into the standard form:

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