# parametrics question (1 Viewer)

#### mathsbrain

##### Member
Hi,

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

Thanks!

#### integral95

##### Well-Known Member
$\bg_white xy = tan^2 \theta - sin^2 \theta$

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

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

#### mathsbrain

##### Member
$\bg_white xy = tan^2 \theta - sin^2 \theta$

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

$\bg_white xy = (\frac{x+y}{2})^2 - (\frac{y-x}{2})^2$
Umm if you try expanding and simplifying don't you get 0=0 which is saying we havent found a catesian relation?

#### integral95

##### Well-Known Member
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 $\bg_white
\frac{sin^2 \theta}{tan^2 \theta}$

Then sub the equations from my 2nd line.

#### fan96

##### 617 pages
Here's another solution:

\bg_white \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.

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

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

The graph of this equation is quite interesting.

Last edited: