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### Focus of a Hyperbola

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Date: 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/
```
Associated Topics:
High School Basic Algebra

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