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Question 895997: Write the standard form equation of the conic section.
Eccentricity = 3/5
Foci: (6,1), (6,-11)
The answer is (x-6)^2 / 64 + (y+5)^2 /100 = 1
I don't understand how to do this problem.
Thank you.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Write the standard form equation of the conic section.
Eccentricity = 3/5
Foci: (6,1), (6,-11)
The answer is (x-6)^2 / 64 + (y+5)^2 /100 = 1
***
Conic section is an ellipse with vertical major axis.(from given foci coordinates)
Its standard form of equation: , a>b, (h,k)=coordinates of center
..
x-coordinate of center=6 (from foci data)
y-coordinate of center=-5 (midpoint between 1 and -11)
center: (6,-5)
c=6(distance from center to foci)
c^2=36
eccentricity=3/5=c/a
3a=5c
a=5c/3=30/3=10
a^2=100
c^2=a^2-b^2
b^2=a^2-c^2=100-36=64
equation:
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