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**Ronald****Guest**

Is x^-1=y a hyperbola's equation? If so what is the a and b in that case?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,017

hi Ronald,

You're thinking of a hyperbola in the format

http://www.mathsisfun.com/geometry/hyperbola.html

y = 1/x is a hyperbola but it doesn't have that format so I cannot give you values for a and b.

The standard format has the x and y axes as the lines of symmetry.

If you rotate the shape 45 degrees you get the format

http://www.mathsisfun.com/sets/function-reciprocal.html

If you want more details of the transformation I could probably work it out (it was a long time ago when I last met this).

I've just read this through to check for mistakes and it's all come back to me.

So transform using

and you get

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Ronald****Guest**

How do you transform a graph 45 degrees?

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,017

hi Ronald,

Usually we have the x axis across and the y axis up, but there is no reason why you shouldn't have a pair of axes in a different position. As long as the two lines are not parallel, you can give the position of points using the new axes. You just need equations that transform the old coordinates into the new ones.

If you look at my diagram below, you'll see I have the ususl axes and I've also marked the new axes in red. The lines are y = x and y = -x

Suppose P is a point on the hyperbola. M is the point below it on the x axis so that x = OM and y = MP.

Draw a line parallel to y = x, from P to N, on the line y = -x.

Q is the point on the x axis so that NQO makes a right angle.

In the new coordinate system ** X = ON** and **Y = NP**

and

I would like to have X and Y as the subject of these two equations.

adding gives

and subtracting gives

Compare this with the transformation I gave in post 2. I seem to have answered your original question.

So using the new coordinate system the equation of the hyperbola becomes

It's the same shape; just described using the new axes.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Ronald****Guest**

It took me some time to understand,but i understood it,thank you.