SOLUTION: I need to Identify the conic section, then place in standard form. The equation is: 4x^2+y^2+8x-4y-28=0 This is as far as i have been able to solve it. (4x^2+8x)+(y^2-4y)=2

Question 141095: I need to Identify the conic section, then place in standard form. The equation is: 4x^2+y^2+8x-4y-28=0
This is as far as i have been able to solve it.
(4x^2+8x)+(y^2-4y)=28
[4(x+1)^2-1]+(y-2)^2-4=28
+4 +4
4(x+1)^2+(y-2)^2-1=32
+1 +1
4(x+1)^2/33+(y-4)^2/33=1

any help you can give would be wonderful. Thank you.
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
4x^2+y^2+8x-4y-28=0

Sort it out and factor the coefficient of x^2. Leave spaces in order to complete the square in the next steps:

To complete the square, you must take half of the x coefficient (Half of 2 is 1, and square which is 1. On the right side, however, you must add 4*1 which is 4. It should look like this:

Divide both sides by 36 to write this in standard form for an ellipse:

This is an ellipse, with center at (-1,2), with a "radius" of 3 in the x direction, and a "radius" of 6 in the y direction.

R^2
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