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
To complete the square for the given equation, 4x^2 + 3y^2 + 8x - 6y - 5 = 0, follow these steps:
1. Rearrange the terms to group the x terms and y terms separately:
4x^2 + 8x + 3y^2 - 6y = 5
2. Factor out the common coefficient from the x terms (4) and y terms (3):
4(x^2 + 2x) + 3(y^2 - 2y) = 5
3. Take half of the coefficient of the x term (2), square it (2^2 = 4), and add it inside the parentheses for the x term. Similarly, take half of the coefficient of the y term (-2), square it (2^2 = 4), and add it inside the parentheses for the y term. Add the corresponding multiples of the squared terms outside the parentheses:
4(x^2 + 2x + 1) + 3(y^2 - 2y + 1) = 5 + 4 + 3
4. Simplify the equation inside the parentheses:
4(x + 1)^2 + 3(y - 1)^2 = 12
5. Divide every term by the constant on the right side (12):
(x + 1)^2/3 + (y - 1)^2/4 = 1
From the final form of the equation, (x + 1)^2/3 + (y - 1)^2/4 = 1, we can see that it represents an ellipse.