Asked by Jordan
How do I complete the square:
4x^2+3y^2+8x-6y-5=0
Also, is this a ellipse or a circle?
first of all arrange the terms that appear to 'belong' together
4x^2 + 8x + 3y^2 - 6y = 5
factor out the 4 from the x terms and the 3 from the y terms
4(x^2 + 2x + ??) + 3(y^2 - 2y + ??)=5
take 1/2 the coefficient of the middle term, square it, then add it
4(x^2 + 2x + 1) + 3(y^2 - 2y + 1) = 5 + 4 + 3
notice I added the 1 inside the bracket which was multiplied by 4, so really I added 4, in the same way for the y terms I added 1 to the inside, but that was multiplied by 3, so really I added 3
that is why on the right side you see a 4 and a 3 added.
now express it in the standard form of an ELLIPSE
4(x+1)^2 + 3(y-1)^2 = 12
and finally divide every term by 12 to get
(x+1)^2 /3 + (y-1)^2 /4 = 1
which I hope you recognize as an ellipse and hope you can read off its properties
4x^2+3y^2+8x-6y-5=0
Also, is this a ellipse or a circle?
first of all arrange the terms that appear to 'belong' together
4x^2 + 8x + 3y^2 - 6y = 5
factor out the 4 from the x terms and the 3 from the y terms
4(x^2 + 2x + ??) + 3(y^2 - 2y + ??)=5
take 1/2 the coefficient of the middle term, square it, then add it
4(x^2 + 2x + 1) + 3(y^2 - 2y + 1) = 5 + 4 + 3
notice I added the 1 inside the bracket which was multiplied by 4, so really I added 4, in the same way for the y terms I added 1 to the inside, but that was multiplied by 3, so really I added 3
that is why on the right side you see a 4 and a 3 added.
now express it in the standard form of an ELLIPSE
4(x+1)^2 + 3(y-1)^2 = 12
and finally divide every term by 12 to get
(x+1)^2 /3 + (y-1)^2 /4 = 1
which I hope you recognize as an ellipse and hope you can read off its properties
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