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.
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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.
Did you graph y=8/x ?
Draw the graph. Is it a hyperbola?
What are the vertices?
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.
Hello, shenton!
Evidently, you've never met this type of hyperbola.Quote:
Determine the distance between the vertices of the hyperbola xy = 8
The graph looks like this:The hyperbola is "tipped" at a 45° angle.Code:
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The x and yaxes 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)
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!