Help wih parametric representation of a function

**Rascot** · June 5th 2012, 10:20 PM

Hi, I'm stuck trying to obtain the cartesian equation of the function defined parametrically by x=1/2(t+1/t), y=1/2(t-1/t).

The book (Mathematics for Engineers, Croft / Davison) gives x^2 - y^2 = 1 as the answer, but I can't figure out how they got there. Ideas?

**earboth** · June 5th 2012, 10:44 PM

Originally Posted by Rascot

Hi, I'm stuck trying to obtain the cartesian equation of the function defined parametrically by x=1/2(t+1/t), y=1/2(t-1/t).

The book (Mathematics for Engineers, Croft / Davison) gives x^2 - y^2 = 1 as the answer, but I can't figure out how they got there. Ideas?

1. Rearrange both equations to:

$2x = t +\frac1t~\wedge~2y=t-\frac1t$

2. Add both sides of the equations:

$2x+2y = 2t~\implies~t = x+y$

3. Replace t in the 1st equation by (x+y):

$x = \frac12 \cdot \left((x+y)+\frac1{x+y} \right)$

4. Expand the brackets and after a few steps of rearranging you'll get the given result.

**Rascot** · June 5th 2012, 11:05 PM

Perfect! Completely forgot about solving simultaneous equations (probably because they are dealt with later in the book). You have my thanks, sir.

By the way, how do you (and pretty much everyone on this forum) post math symbols as images? Is there a piece of software you guys use?

**earboth** · June 5th 2012, 11:26 PM

Originally Posted by Rascot

Perfect! Completely forgot about solving simultaneous equations (probably because they are dealt with later in the book). You have my thanks, sir.

By the way, how do you (and pretty much everyone on this forum) post math symbols as images? Is there a piece of software you guys use?

1. Go to LaTeX Help Forum. You'll find 2 tutorials where you can find how to use the implemented Latex-compiler.

2. If you need the synatx of a Latex command you'll find it probably here: Help

isplaying a formula - Wikipedia, the free encyclopedia

**Soroban** · June 6th 2012, 05:30 AM

Hello, Rascot!

Hi, I'm stuck trying to obtain the cartesian equation of the function
. . defined parametrically by: . $\begin{Bmatrix}x &=& \frac{1}{2}(t+\frac{1}{t}) \\ \\[-3mm] y&=& \frac{1}{2}(t-\frac{1}{t}) \end{Bmatrix}$

The book (Mathematics for Engineers, Croft/Davison) gives $x^2 - y^2 \,=\, 1$ as the answer.

Square the equations: . $\begin{Bmatrix}x^2 &=& \frac{t^2}{4} + \frac{1}{2} + \frac{1}{4t^2} & [1] \\ \\[-4mm] y^2 &=& \frac{t^2}{4} - \frac{1}{2} + \frac{1}{4t^2} & [2] \end{Bmatrix}$

Subtract [2] from [1]: . . $x^2 - y^2 \;=\;1$

**Rascot** · June 6th 2012, 11:59 AM

Thank you both very much for your help. I can't mark the thread as solved, but would kindly ask moderators to do so if they happen to come across it.

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