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Question 84656:
Answer by faith74(1)   (Show Source):
You can put this solution on YOUR website! Write the equation in standard form and classify the conic section 3x^2+2y^2+24x+20y+92=0
You must complete the square
first for the x's
3x^2 + 24x = -92
3(x^2 + 8x + 16) = -92 + 48
3(x + 4)^2 = -44
Now for the y's (you must use the constant that you left off with on the x's)
2y^2 + 20y = -44
2(y^2 + 10y + 25) = -44 + 50
2(y + 5)^2 = 6
Put them together.
3(x + 4)^2 + 2(y + 5)^2 = 6
Divide by the constant (6)
[(x + 4)^2]/2 + [(y + 5)^2]/3 = 1
Since the denominators are different and it is addition, this means it is an ellipse (oval) with center (-4,-5) with horizontal distance from center being the square root of 2 and the vertical distance from center being the square root of 3.
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