SOLUTION: The equation of an ellipse is 16x^2+25y^2-128x-150y+381=0. Determine the following: a.) coordinates of the center of the ellipse b.) length of the minor axis c.) distance betwee

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The equation of an ellipse is 16x^2+25y^2-128x-150y+381=0. Determine the following: a.) coordinates of the center of the ellipse b.) length of the minor axis c.) distance betwee      Log On


   

Question 953568: The equation of an ellipse is 16x^2+25y^2-128x-150y+381=0. Determine the following:
a.) coordinates of the center of the ellipse
b.) length of the minor axis
c.) distance between foci
Answer by lwsshak3(11628) About Me  (Show Source):
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The equation of an ellipse is 16x^2+25y^2-128x-150y+381=0. Determine the following:
a.) coordinates of the center of the ellipse
b.) length of the minor axis
c.) distance between foci
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16x^2+25y^2-128x-150y+381=0
16x^2-128x+25y^2-150y=-381
complete the squares:
16(x^2-8x+16)+25(y^2-6y+9)=-381+256+225
16(x-4)^2+25(y-3)^2=100
divide both sides by 100
%28x-4%29%5E2%2F%28100%2F16%29%2B%28y-3%29%5E2%2F4=1
%28x-4%29%5E2%2F%2825%2F4%29%2B%28y-3%29%5E2%2F4=1
ellipse has a horizontal major axis.(larger denominator under x-term)
Its standard form of equation: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1, a>b, (h,k)=coordinates of center
a.)center: (4,3)
a^2=25/4
b^2=4
b=2
b.)length of minor axis=2b=4
foci:
c^2=a^2-b^2=(25/4)-4=25/4-16/4=9/4
c=3/2
c.)distance between foci=2c=3