# parametrics question

• 16 Mar 2019, 12:43 AM
mathsbrain
parametrics question
Hi,

Stuck on this question:
convert x=tanx-sinx and y=tanx+sinx to cartesian form.

Thanks!
• 16 Mar 2019, 9:54 AM
integral95
Re: parametrics question
$xy = tan^2 \theta - sin^2 \theta$

$x + y = 2tan \theta \ \ \ y-x = 2sin \theta$

$xy = (\frac{x+y}{2})^2 - (\frac{y-x}{2})^2$
• 16 Mar 2019, 1:14 PM
mathsbrain
Re: parametrics question
Quote:

Originally Posted by integral95
$xy = tan^2 \theta - sin^2 \theta$

$x + y = 2tan \theta \ \ \ y-x = 2sin \theta$

$xy = (\frac{x+y}{2})^2 - (\frac{y-x}{2})^2$

Umm if you try expanding and simplifying don't you get 0=0 which is saying we havent found a catesian relation?
• 16 Mar 2019, 8:10 PM
integral95
Re: parametrics question
Quote:

Originally Posted by mathsbrain
Umm if you try expanding and simplifying don't you get 0=0 which is saying we havent found a catesian relation?

True that, you should change tan^2 to sin^2/cos^2 then simplify to get $
\frac{sin^2 \theta}{tan^2 \theta}$

Then sub the equations from my 2nd line.
• 18 Mar 2019, 1:03 AM
fan96
Re: parametrics question
Here's another solution:

\begin{aligned} \frac{y-x}{x+y} &= \cos \theta \\ 1 - \left(\frac{y-x}{x+y}\right)^2 &=\left(\frac{y-x}{2}\right)^2 \\ (x+y)^2- (y-x)^2 &= \frac 1 4 (y-x)^2(x+y)^2 \\ 16xy &= (x^2-y^2)^2 \end{aligned}

i.e.

$(x^2-y^2)^2-16xy=0$

It's also clear to see solving this equation with $x + y = 0$ gives the point $(0, 0)$ just as it would for the parametric form. Therefore this is a valid conversion.

The graph of this equation is quite interesting.