How do you write an equation for a ellipse with center (5, -4), vertical major axis of length 12, and minor axis of length 8?

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How do you write an equation for a ellipse with center (5, -4), vertical major axis of length 12, and minor axis of length 8?

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Answer 1 · 2016-06-05T02:17:14

$\frac{{(x - 5)}^{2}}{36} + \frac{{(y + 4)}^{2}}{16} = 1$ $(x - 5)^2/36 + (y + 4)^2/16 = 1$

Explanation:

Standard equations of an ellipse

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

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

where:
Center: $(h, k)$ $(h, k)$
Major axis: $2 a$ $2a$
minor axis: $2 b$ $2b$

In the given
Center: $(5, - 4)$ $(5, -4)$
Major axis: $12 = 2 a \Rightarrow a = 6$ $12 = 2a => a = 6$
minor axis: $8 = 2 b \Rightarrow b = 4$ $8 = 2b => b = 4$

We are dealing with an ellipse with a vertical major axis, so we should use the second form of the standard equation

$\frac{{(x - 5)}^{2}}{6^{2}} + \frac{{(y + 4)}^{2}}{4^{2}} = 1$ $(x - 5)^2/6^2 + (y + 4)^2/4^2 = 1$

$\Rightarrow \frac{{(x - 5)}^{2}}{36} + \frac{{(y + 4)}^{2}}{16} = 1$ $=> (x - 5)^2/36 + (y + 4)^2/16 = 1$

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How do you write an equation for a ellipse with center (5, -4), vertical major axis of length 12, and minor axis of length 8?

1 Answer

Explanation:

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