parametrics question (1 Viewer)

mathsbrain · Mar 16, 2019

Hi,

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

Thanks!

integral95 · Mar 16, 2019

$xy = tan^2 \theta - sin^2 \theta$

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

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

mathsbrain · Mar 16, 2019

integral95 said:
$xy = tan^2 \theta - sin^2 \theta$

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

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

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

integral95 · Mar 16, 2019

mathsbrain said:
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.

fan96 · Mar 18, 2019

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.

parametrics question (1 Viewer)

mathsbrain

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integral95

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mathsbrain

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integral95

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fan96

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