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What is the equation of the ellipse with foci at (5, 0), (-5, 0) and vertices (9, 0), (-9, 0)?

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What is the equation of the ellipse with foci at (5, 0), (-5, 0) and vertices (9, 0), (-9, 0)?
A. x squared divided by 81 plus y squared divided by 56 equals 1
B. x squared divided by 81 minus y squared divided by 56 equals 1
C. x squared divided by 81 plus y squared divided by 25 equals 1
D. x squared divided by 81 minus y squared divided by 25 equals 1

 

asked Sep 9, 2014 in ALGEBRA 2 by tonymate Pupil

1 Answer

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Best answer

The vertices of the ellipse are (9, 0) and (-9, 0) and its foci are (5, 0) and (-5, 0).

The vertices and foci are respectively a and c units from the center (hk) and the relation between ab and c is b2 = a2 - c2.

The center (h, k) = [ (x₁ + x₂)/2, (y₁ + y₂)/2 ] = [ (9 - 9)/2, (0 + 0)/2 ] = (0, 0).

-----------------------------------------

To find the value of a, 

2a = distance between vertices

2a = sqrt((0-(0))²+(9-(-9))²)

    = 18

a = 9

To find the value of c, 

2c = distance between foci

2c = sqrt((0-0)²+(5-(-5))²)

    = 10

c=5

--------------------------------------------

To find the value of b, 

b² = a2 - c².

b² = 81 - 25

b² =56

The equation of the hyperbola is

x2/81 + y2/56 = 1.

So option A is correct.

answered Sep 9, 2014 by bradely Mentor
edited Sep 9, 2014 by moderator

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