Asked by Anonymous
Find the equation of the ellipse with the center at (2,-3), one focus at (3,-3), and one vertex at (5,-3). Also graph the equation.
Answers
Answered by
Steve
the center says that the ellipse is
(x-2)^2/a^2 + (y+3)^2/b^2 = 1
Since both vertex and focus are at y=-3, the long axis is horizontal.
5-2=3, so a=3
3-2=1, so c=1
b^2+c^2 = a^2, so b^2 = 8
(x-2)^2/9 + (y+3)^2/8 = 1
graph and analysis at
http://www.wolframalpha.com/input/?i=focus%2C+vertex+for+%28x-2%29^2%2F9+%2B+%28y%2B3%29^2%2F8+%3D+1
(x-2)^2/a^2 + (y+3)^2/b^2 = 1
Since both vertex and focus are at y=-3, the long axis is horizontal.
5-2=3, so a=3
3-2=1, so c=1
b^2+c^2 = a^2, so b^2 = 8
(x-2)^2/9 + (y+3)^2/8 = 1
graph and analysis at
http://www.wolframalpha.com/input/?i=focus%2C+vertex+for+%28x-2%29^2%2F9+%2B+%28y%2B3%29^2%2F8+%3D+1
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