How do I find the center of an ellipse in standard form?

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Answer 1 · 2014-10-19T02:40:35

An ellipse in standard form looks like

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

or

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

where $a \geq b$ $a >= b$ .
However, for your question, we don't need to be concerned with $a$ $a$ and $b$ $b$

An ellipse in standard form is centered at (h, k).

For example,

$\frac{{(x - 1)}^{2}}{1} + \frac{{(y + 2)}^{2}}{4} = 1$ $(x - 1)^2/1 + (y + 2)^2/4 = 1$ is centered at (1, -2)

$\frac{x^{2}}{16} + \frac{{(y - 5)}^{2}}{9} = 1$ $x^2/16 + (y - 5)^2/9 = 1$ is centered at (0, 5)