Focus of a HyperbolaDate: 04/06/98 at 01:44:13 From: Saritha Komatireddy Subject: Algebra 2 In a hyperbola, let the x-radius be a and the y-radius be b. Then a squared + b squared = c squared (focal radius), right? At least, that's what it says in the math book. Well, what I'm asking is, how come when you rotate the focus point around the center (as if on the edge of a circle), it happens to hit the asymptote with the same x coordinate as the vertex? How come it makes that perfect right triangle? Is there a proof you can show me? Date: 04/06/98 at 08:28:58 From: Doctor Rob Subject: Re: Algebra 2 The equation of the hyperbola is x^2/a^2 - y^2/b^2 = 1. The asymptotes are the lines y = (b/a)*x and y = -(b/a)*x. The vertices are (a,0) and (-a,0). The foci are (c,0) and (-c,0). The hyperbola is defined as the locus of points such that the absolute value of the difference between the distances from the point to the two foci is a constant. By considering one of the two vertices, you can see that the constant is 2*a. Now consider the point (sqrt[2]*a, b). Its distance from (c,0) is: sqrt[(sqrt[2]*a-c)^2+b^2] and its distance from (-c,0) is: sqrt[(sqrt[2]*a+c)^2+b^2]. The difference of these distances must be 2*a, so: sqrt[(sqrt[2]*a-c)^2+b^2] - sqrt[(sqrt[2]*a+c)^2+b^2]] = 2*a When you simplify this equation, you find that: c^2 = a^2 + b^2 This means that the distance from (0,0) to (a,b) (a point on the asymptote directly above the vertex (a,0)) is c, as you stated above. I hope this is what you had in mind as a proof. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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