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Question 259713: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find.
a. The Center C
b. Length of Major Axis
c. Length of Minor Axis
d. Distance from C to foci
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0
Complete the square on the x-terms and on the y-terms:
2(x^2+16x+?) + 6(y^2-8y+?) = -212+?
2(x^2+16x+64) + 6(y^2-8y+16) = -212+2*64+6*16
2(x+8)^2 + 6(y-4) = 12
Divide thru by 12 to get:
(x+8)^2/6 + (y-4)^2/2 = 1
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h = -8 ; k = 4
a = sqrt(6) ; b = sqrt(2)
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find.
a. The Center C ..........(h,k)
b. Length of Major Axis...2a
c. Length of Minor Axis,,,2b
d. Distance from C to foci:
a^2 + b^2 = distance^2
d = sqrt(6 + 2)
d = 2sqrt(2)
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Cheers,
Stan H.
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