Question 454235: Formula for an ellipse= (x+1)^24(y+3)^2=1
what are the vertices and foci of this ellipse
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Formula for an ellipse= (x+1)^24(y+3)^2=1
what are the vertices and foci of this ellipse
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Do you mean (x+1)^2 + 4(y+3)^2=1 ?
(x+1)^2/1 + (y+3)^2/0.25 = 1
a^2 = 1, b^2 = 1/4
Center at (-1,-3), long axis parallel to the x-axis
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Distance to foci = sqrt(a^2 - b^2) = sqrt(0.75)
Foci at (-1-sqrt(0/75),-3) and (-1+sqrt(0.75,-3)
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Vertices 1 unit from center parallel to x-axis and 1/2 unit vertical
--> (-2,-3) and (0,-3)
and (-1,-3.5) and (-1,-2.5)
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