How do you write an equation of an ellipse in standard form given focus at (0,0) and vertices at (2, pi/2) and (8, 3pi/2)?

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How do you write an equation of an ellipse in standard form given focus at (0,0) and vertices at (2, pi/2) and (8, 3pi/2)?

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Answer 1 · 2016-02-18T09:05:43

$\frac{x^{2}}{16} + \frac{{(y + 3)}^{2}}{25} = 1$ $x^2/16+(y+3)^2/25=1$

Explanation:

The center of the ellipse can be solved by obtaining the average value of the vertices at $(2, \frac{π}{2}) = (0, 2)$ $(2, pi/2)=(0, 2)$ and $(8, \frac{3 π}{2}) = (0, - 8)$ $(8, (3pi)/2)=(0, -8)$

Center $(h, k) = (0, - 3)$ $(h, k)=(0, -3)$

there is a focus at (0, 0), vertex at (0, 2) and center at (0, -3) so that $c = 3$ $c=3$ and $a = 5$ $a=5$ by computation.

solve $b$ $b$ :

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

$5^{2} = b^{2} + 3^{2}$ $5^2=b^2+3^2$

$b^{2} = 25 - 9$ $b^2=25-9$
$b^{2} = 16$ $b^2=16$
and $b = 4$ $b=4$

the equation of the ellipse with vertical major axis is:

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

$\frac{{(x - 0)}^{2}}{4^{2}} + \frac{{(y - - 3)}^{2}}{5^{2}} = 1$ $(x-0)^2/4^2+(y--3)^2/5^2=1$

$\frac{x^{2}}{16} + \frac{{(y + 3)}^{2}}{25} = 1$ $x^2/16+(y+3)^2/25=1$

see the graph

graph{x^2/16+(y+3)^2/25=1[-20, 20,-10,10]}

have a nice day! from the Philippines..

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How do you write an equation of an ellipse in standard form given focus at (0,0) and vertices at (2, pi/2) and (8, 3pi/2)?

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Explanation:

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