Hyperbola - Why is c^2 = a^2 + b^2?

[masters]

Ok, you'll have to set this one up yourself because I can't seem to copy and paste my figure into this space:

Hyperbola opens left and right.
Center (0,0)
foci (-c,0), (+c,0)
Vertices (-a, 0), (+a,0)
The transverse axis is 2a and the conjugate axis is 2b

The distance from the center to the endpoint on the conjugate axis is b.
The distance from the center to a vertex is a.

The distance from a vertex to the endpoint of the conjugate axis (hypotenuse of the right triangle) is sqrt (a[sup:2s7xz8e9]2[/sup:2s7xz8e9]+ b[sup:2s7xz8e9]2[/sup:2s7xz8e9]) which is somehow equal to the distance from the center to the focus which is c. How is that? How do I know that c[sup:2s7xz8e9]2[/sup:2s7xz8e9] = a[sup:2s7xz8e9]2[/sup:2s7xz8e9] + b[sup:2s7xz8e9]2[/sup:2s7xz8e9] in this context?

Hope you can either draw the picture or produce one for reference. Thanks. Dale

 

[Denis]

Look for right triangle(s) with legs a and b. OK?

 

[masters]

I understand where the right triangle is and one leg is a and the other is b. I understand those lengths. That makes the hypotenuse sqrt(a[sup:2jrn9cmy]2[/sup:2jrn9cmy] + b[sup:2jrn9cmy]2[/sup:2jrn9cmy]). My question is why is that hypotenuse the same length as the distance from the focus to the vertex?

In the figure V[sub:2jrn9cmy]1[/sub:2jrn9cmy]B[sub:2jrn9cmy]1[/sub:2jrn9cmy] = sqrt(a[sup:2jrn9cmy]2[/sup:2jrn9cmy] + b[sup:2jrn9cmy]2[/sup:2jrn9cmy]).

Why is that equal to the distance from the focus to the center (0,0)?

 

[Deleted member 4993]

What is the definition of focus (mathematical) of a hyperbola?

It is by definition c = sqrt(a^2 + b^2)

If you have that - then you can show that the difference of distances from each focus of any point on the hyperbola remains constant.(definition of hyperbola)

It is kind of bass-ackwards, but that's the way it is!!!