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On 5/31/01 7:57:00 PM, Anonymous wrote:
>i need help on these problems... First check to see whether you have an x² term and a y² term. If one of them is missing, it's a parabola. Then, assuming it's not a parabola, put every term on one side of the equation and zero on the other side. Complete all the squares, and then put whatever constant is left over on the other side of the equation. When you do that you'll be left with a(x-h)² + b(y-k)² = c If c is zero, you have no solution, a point or a pair of intersecting lines, and you're done. Otherwise, change all the signs if necessary so c is positive. If both a and b are negative, then there is no solution. If both a and b are positive, it's an ellipse. An ellipse with a=b is also a circle. If a and b have opposite signs, it's a hyperbola. OK, now I'll do your example.
4x²=y²+8y+32 So it's a hyperbola because there's an x² term and a y² term, and a and b have opposite signs. Related pages in this website
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