How do you write an equation of a ellipse with foci (0,0), (0,8) and major axis of length 16?

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How do you write an equation of a ellipse with foci (0,0), (0,8) and major axis of length 16?

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Answer 1 · 2016-12-30T10:03:05

$\frac{x^{2}}{48} + \frac{{(y - 4)}^{2}}{64} = 1$ $x^2/48 + (y - 4)^2/64 = 1$

Explanation:

The midpoint between the foci is the center

$C : (\frac{0 + 0}{2}, \frac{0 + 8}{2})$ $C: ((0 + 0)/2, (0 + 8)/2)$

$\Rightarrow C : (0, 4)$ $=> C: (0, 4)$

The distance between the foci is equal to $2 c$ $2c$

$2 c = \sqrt{{(0 - 0)}^{2} + {(0 - 8)}^{2}}$ $2c = sqrt((0 - 0)^2 + (0 - 8)^2)$

$\Rightarrow 2 c = \sqrt{0 + 64}$ $=> 2c = sqrt(0 + 64)$

$\Rightarrow 2 c = 8$ $=> 2c = 8$

$\Rightarrow c = 4$ $=> c = 4$

The major axis length is equal to $2 a$ $2a$

$2 a = 16$ $2a = 16$

$\Rightarrow a = 8$ $=> a = 8$

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

$\Rightarrow b^{2} = a^{2} - c^{2}$ $=> b^2 = a^2 - c^2$

$\Rightarrow b^{2} = 8^{2} - 4^{2}$ $=> b^2 = 8^2 - 4^2$

$\Rightarrow b^{2} = 64 - 16$ $=> b^2 = 64 - 16$

$\Rightarrow b^{2} = 48$ $=> b^2 = 48$

Between the coordinates of the foci, only the y-coordinate changes, this means the major axis is vertical. The standard equation of an ellipse with a vertical major axis is

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

$\Rightarrow \frac{{(x - 0)}^{2}}{48} + \frac{{(y - 4)}^{2}}{8^{2}} = 1$ $=> (x - 0)^2/48 + (y - 4)^2/8^2 = 1$

$\Rightarrow \frac{x^{2}}{48} + \frac{{(y - 4)}^{2}}{64} = 1$ $=> x^2/48 + (y - 4)^2/64 = 1$

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How do you write an equation of a ellipse with foci (0,0), (0,8) and major axis of length 16?

1 Answer

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