# Distance between vertices

• Mar 19th 2007, 08:02 AM
shenton
Distance between vertices
Determine the distance between the vertices of the hyperbola xy=8.

I don't know how to do this one as it is a strange one. What happened to the x² and y²?

Thanks.
• Mar 19th 2007, 08:35 AM
Plato
Did you graph y=8/x ?
Draw the graph. Is it a hyperbola?
What are the vertices?
• Mar 19th 2007, 08:34 PM
shenton
I believe a hyperbola opens either up/down or left/right. Y=8/x does not opens in this manner, it opens up like a reciprocal function.

I'm not sure - this question comes under the hyperbola section. I'm suppose to calculate the vertices algebratically.
• Mar 19th 2007, 08:56 PM
Soroban
Hello, shenton!

Quote:

Determine the distance between the vertices of the hyperbola xy = 8
Evidently, you've never met this type of hyperbola.

The graph looks like this:
Code:

```                          |                           |*                           |                           | *                           |  *                           |    *                           |        *                           |              *       - - - - - - - - - - + - - - - - - - - - -           *              |                 *        |                     *    |                       *  |                         * |                           |                         *|                           |```
The hyperbola is "tipped" at a 45° angle.
The x- and y-axes are the asymptotes.

The vertices lie on the line y = x.
So we have: .x² = 8 . . x = ±2√2
. . . . . . . . . . . . . . . _ . . ._ . . . - . . ._ . . . _
The vertices are: .(2√2, 2√2) and (-2√2, -2√2)

• Mar 20th 2007, 12:32 AM
shenton
Thanks for the detailed explaination. Yes, I have never seen this type of hyperbola.

I think I can work out the distance now.

Given vertices (2√2, 2√2) and (-2√2, -2√2) of hyperbola xy=8:

The distance is:

√ [ (-2√2 - 2√2)² + (-2√2 - 2√2)² ]

= √ [ (-4√2) ² + (-4√2)² ]

= √ (32 + 32)

= √ 64

= 8

Thanks!