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Write an equation of the ellipse with foci at (0, ±10), and vertices at (0, ±11).Asked by bear
write an equation of the ellipse with foci at (0,+11) and verticies at (0,+12)
Answers
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
Since when does an ellipse have its center midway between its focus and its vertex?
Of course, the problem was poorly written. It should have read
write an equation of the ellipse with foci at (0,±11) and vertices at (0,±12)
Then we have
a = 12
c = 11
b^2 = 23
and the equation is
x^2/23 + y^2/144 = 1
Since when does an ellipse have its center midway between its focus and its vertex?
Of course, the problem was poorly written. It should have read
write an equation of the ellipse with foci at (0,±11) and vertices at (0,±12)
Then we have
a = 12
c = 11
b^2 = 23
and the equation is
x^2/23 + y^2/144 = 1
Answered by
Bot
The center of the ellipse is halfway between the foci and is at (0,11.5). The distance from the center to the vertex is a=0.5. The distance from the center to each focus is c=0.5.
The equation of the ellipse is:
(x-0)^2/0.25 + (y-11.5)^2/1 = 1
Simplified:
x^2/0.25 + (y-11.5)^2 = 1
The equation of the ellipse is:
(x-0)^2/0.25 + (y-11.5)^2/1 = 1
Simplified:
x^2/0.25 + (y-11.5)^2 = 1
Answered by
Bot
My apologies for the mistake. Thank you for correcting it. You are correct, the center of the ellipse is at (0,0) and the semi-major axis is 12. The semi-minor axis can be found using the formula b^2 = c^2 - a^2, where c is the distance from the center to the focus. So, b^2 = 11^2 - 12^2 = 23.
Therefore, the equation of the ellipse is
x^2/23 + y^2/144 = 1.
Therefore, the equation of the ellipse is
x^2/23 + y^2/144 = 1.