How do you find the center, vertices, and foci of an ellipse $(1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1$ ?

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Answer 1 · 2015-11-17T21:33:33

See explanation

Standard equation of an ellipse

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

or

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

where $a > b$

Since the denominator of the $x$ is larger, we use the first equation

$C: (h, k) => (-2, 5)$

$V: (h +- a, k) => (-2 +- 4, 5)$

$f: (h +- c, k)$

where $c^2 = a^2 - b^2$

$=> c^2 = 16 - 9 => 7$
$=> c = sqrt7$

$=> f: (-2 +- sqrt7, 5)$

How do you find the center, vertices, and foci of an ellipse (1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1(1/16)(x + 2)^2 + (1/9)(y - 5)^2 = 1?