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Question 383818: Directions: For each of the following conic sections identify whether it is a circle, parabola, ellipse or hyperbola. Find its center and/or vertices, its focus or foci, major or transverse axis, if appropriate, asymptotes or directrix if present. Graph each of them, labeling all important points.
The question I'm having trouble with is:
4x^2+y^2-24x+4y+36=0
So far, I've gotten this, but I think I did something wrong:
4x^2+y^2-24x+4y+36=0
[-4(x+3)^2-9]+y^2+4y+36=0
[-4(x+3)^2-9]+(y+2)^2-4+36=0
[-4(x+3)^2]+(y+2)^2=-68
*divide by -68*
[((x+3)^2)/7] + [((y+2)^2)/-68]=1
I think it's an ellipse, but I'm not sure. The -68 is really throwing me off.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The -68 should throw you off. That's because your arithmetic is faulty. And your arithmetic is faulty because you are not following the standard process.
The first thing is to make the identification of the type of conic.
If you have an  term but no  term (or vice versa), then you have a parabola.
If you have opposite signs on the and terms, then you have a hyperbola.
If you have the same signs, but different coefficients (as in your example) you have an ellipse. The squared variable with the smaller coefficient identifies the major axis.
Otherwise it is a circle.
Both variables squared (not parabola), same sign on squared variables (not hyperbola), different coefficients on the squared variables (not circle, therefore ellipse)
The first step is to move the constant term to the RHS and arrange like variable terms together:
Start with the  terms. If there is a lead coefficient, then factor it out of the terms:
\ +\ y^2\ +\ 4y\ =\ -36)
Divide the coefficient on the first degree term by 2, square the result, add that result INSIDE the parentheses, then add the same number TIMES the lead coefficient to the RHS: (-6 divided by 2 is -3, -3 squared is 9, put 9 inside the parentheses, then add 4 times 9 = 36 to the RHS)
\ +\ y^2\ +\ 4y\ =\ -36\ +\ 36)
Factor the perfect square in :
^2\ +\ y^2\ +\ 4y\ =\ -36\ +\ 36)
No lead coefficient on the  terms, so just divide the first degreee term by 2, square the result, add that result to both sides:
^2\ +\ y^2\ +\ 4y\ +\ 4\ =\ -36\ +\ 36\ +\ 4)
Factor the terms and collect terms in the RHS.
^2\ +\ \left(y\ +\ 2\right)^2\ =\ 4)
Divide through by the constant in the RHS:
^2\ +\ \frac{\left(y\ +\ 2\right)^2}{4}\ =\ 1)
For the purposes of easy pattern matching I'm going to use the following equivalent equation:
^2}{1^2}\ +\ \frac{\left(y\ -\ (-2)\right)^2}{2^2}\ =\ 1)
Consider the canonical equation for an ellipse with a vertical major axis:
^2}{b^2}\ +\ \frac{\left(y\ -\ k\right)^2}{a^2}\ =\ 1) .
which has a center at ) , vertices at ) and ) , foci at ) and ) -- where  , the end points of the major axis at ) and ) . The directrices are the lines  and  .
The only value you need to calculate from your standard form equation is  since the others are available by inspection.  . Plug in the numbers and you will have all of the answers you need. If you write back, I'll send you a nice looking graph with all of the labels, etc.
John
My calculator said it, I believe it, that settles it
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