Deriving the Hyperbola FormulaDate: 04/27/98 at 17:23:07 From: Gauteaux Subject: Hyperbolas When speaking of hyperbolas, why does: C squared = A squared + B squared? My entire math class, including my teacher, has tried to figure this out, but no one can come up with a logical reason. My teacher said that if we could come up with an explanation of this, we would get an A on our next test. Can you please help? Date: 04/28/98 at 17:12:27 From: Doctor Jerry Subject: Re: Hyperbolas Hi Gauteaux, I wouldn't want to say that there is one reason everyone would accept, but here's a reason many would accept. A hyperbola can be defined as follows (ellipses have a similar definition): Two points, called foci, are given; they are 2c units apart. A hyperbola is the locus of all points for which the difference of the distances to the foci is a constant 2a, where 2a > 2c. To take a special case, suppose the two foci are at (-c,0) and (c,0). Then if (x,y) is on the ellipse, we must have: sqrt((x-c)^2+y^2) - sqrt((x+c)^2+y^2) = 2a This is the equation of one "branch." To keep things simple, I'll stick with this case. The other case is similar. This equation can be greatly simplified. First, write the equation as: sqrt((x-c)^2+y^2) = sqrt((x+c)^2+y^2) + 2a Square both sides and simplify. You will get: -c*x-a^2 = a*sqrt((x+c)^2+y^2) Again, square both sides and simplify. You will get: (c^2-a^2)x^2 - a^2*y^2 = a^2(c^2 - a^2) It is common to set c^2-a^2 = b^2, to simplify the equation. b^2*x^2-a^2*y^2 = a^2*b^2 x^2/a^2 - y^2/b^2 = 1. Fortunately, the constant b is interesting. If you draw about the origin a rectangle that is 2a by 2b and then draw its diagonals, the diagonals are the asymptotes of the hyperbola. Please check my algebra. I can make mistakes. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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