How do you write an equation of an ellipse in standard form given center (-1, 3) vertex (3,3) and minor axis of length 2?

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How do you write an equation of an ellipse in standard form given center (-1, 3) vertex (3,3) and minor axis of length 2?

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Answer 1 · 2016-01-26T15:41:09

$\frac{{(x + 1)}^{2}}{16} + \frac{{(y - 3)}^{2}}{4} = 1$ $(x+1)^2/16 + (y-3)^2/4 = 1$

Explanation:

The standard form for an ellipse is
$\frac{{(x - h)}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^"/a^2 +(y-k)^2/b^2 = 1$
where $(h, k)$ $(h,k)$ is the centre of the ellipse, $a$ $a$ is the distance from the centre to the vertices and $c$ $c$ is the distance from the centre to the foci. $b$ $b$ is the minor axis.

$b^{2} + c^{2} = a^{2}$ $b^2+c^2 = a^2$

In this example $a = 3 - (- 1) = 4$ $a = 3 - (-1) = 4$ (The difference if the $x$ $x$ coordinates of the centre and the vertex.)

The equations is $\frac{{(x + 1)}^{2}}{16} + \frac{{(y - 3)}^{2}}{4} = 1$ $(x+1)^2/16 + (y-3)^2/4 = 1$

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How do you write an equation of an ellipse in standard form given center (-1, 3) vertex (3,3) and minor axis of length 2?

1 Answer

Explanation:

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