SOLUTION: Find the center, vertices, and foci of the ellipse with equation x squared divided by 144 plus y squared divided by 225 = 1. (5 points) Center: (0, 0); Vertices: (0, -15), (0,

Algebra ->  Trigonometry-basics -> SOLUTION: Find the center, vertices, and foci of the ellipse with equation x squared divided by 144 plus y squared divided by 225 = 1. (5 points) Center: (0, 0); Vertices: (0, -15), (0,       Log On


   

Question 1138080: Find the center, vertices, and foci of the ellipse with equation x squared divided by 144 plus y squared divided by 225 = 1. (5 points)

Center: (0, 0); Vertices: (0, -15), (0, 15); Foci: (-12, 0), (12, 0)

Center: (0, 0); Vertices: (0, -15), (0, 15); Foci: (0, -9), (0, 9)

Center: (0, 0); Vertices: (-15, 0), (15, 0); Foci: (0, -12), (0, 12) )

Center: (0, 0); Vertices: (-15, 0), (15, 0); Foci: (-9, 0), (9, 0)
Answer by MathLover1(20850) About Me  (Show Source):
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Find the center, vertices, and foci of the ellipse with equation
+x%5E2%2F144+%2B+y%5E2%2F225+=+1
compare to standard formula: +%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1
so, your ellipse is centered at origin; h=0,k=0
also,
a%5E2=144=>a=12
b%5E2=225=>b=15
c+= distance to focus
For an ellipse with major axis parallel to the y-axis, the Foci (focus points ) are defined as :
(h,+k%2Bc ), (h, k-c ), where c=+sqrt%28b%5E2-a%5E2%29 is the distance from the center : (h, k+) to a focus

c=+sqrt%28b%5E2-a%5E2%29
c=+sqrt%28225-144%29
c=sqrt%2881%29

c=9
so, a focus will be at
(0,+0%2B9 ), (0, 0-9 )
(0,+9 ), (0, -9 )
the vertices are at (h,+k%2Bb ), (h, k-b )
we have b=15
the vertices are at (0,+15 ), (0, -15 )

answer: Center: (0, 0); Vertices: (0, -15), (0, 15); Foci: (0,+-9), (0, 9)