SOLUTION: Focus at (0,1); 4/3 is One-Half the Length of Minor Axis. Write equation of the ellipse that has center at the Origin.

Question 539898:
Focus at (0,1); 4/3 is One-Half the Length of Minor Axis.
Write equation of the ellipse that has center at the Origin.
Found 2 solutions by lwsshak3, acolunga3:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Focus at (0,1); 4/3 is One-Half the Length of Minor Axis.
Write equation of the ellipse that has center at the Origin.
**
This is an ellipse with vertical major axis. Standard form of equation:
(x-h)^2/b^2-(y-k)^2/a^2=1, a>b, (h,k) being (x,y) coordinates of the center.
center:(0,0)
One-Half the Length of Minor Axis=4/3=b
b^2=16/9
c=distance from center to focus=1
c^2=1^1=1
c^2=a^2-b^2
a^2=c^2+b^2=1+16/9=9/9+16/9=25/9
Equation of given ellipse:
9x^2/16+9y^2/25=1
Answer by acolunga3(2) (Show Source): You can put this solution on YOUR website!
I worked with some friends:
Origin: Center: (0,0)
Focus: 0,1
4/3 is 1/2 length of Minor Axis SO 4/3 * 1/2 = 16/9

first mark: 0,0 as h,k
c^2 = a^2 / a^2 - b^2 / b^2 =1

C^2 = A^2 - B^2
1^2 = a^2 - 4/3 ^2 => 1=a^2 - 16/9 so
1 + 16/9 =a^2 => 9/9 + 16/9 => a^2 =25/9

x^2 / 16/9 b^2 + y^2 / 25/9 a^2 =1 now multiply 9th fraction to top and bottom of the fraction to get
9x^2/16 + 9y^2/25 =1 as the answer

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