Parametric Trig Functions

Question

A closed curve in the $(x, y)$-plane is represented by the functions

$$x(θ)=\frac12(\cos \theta +\sqrt2 (\sin \theta))$$

$$y(θ)=\frac12(− \cos \theta +\sqrt2 (\sin \theta))$$

where the parameter $\theta$ lies in the range $0 \leqslant \theta < 2\pi$.

Using the identity $\cos^2\theta + \sin^2 \theta = 1$, or otherwise, show that the curve may be expressed as the set of points $(x, y)$ satisfying $3x^2 − 2xy + 3y^2 = 2$.

Answer 1

Clearly

$1) \to$ $$2x=\cos(\theta)+\sqrt{2}\sin(\theta)$$

$2) \to $ $$2y=-\cos(\theta)+\sqrt{2}\sin(\theta)$$

Thus by $1)+2)$ we get ,

$$2x+2y=2\sqrt{2}\sin(\theta) $$

So $$\frac{x+y}{\sqrt{2}}=\sin(\theta)$$

Thus by $1)-2)$ we get ,

$$2x-2y=2\cos(\theta) $$

So $$x-y=\cos(\theta)$$

Now clearly $$ \sin^2(\theta)+\cos^2(\theta)= \left(\frac{x+y}{\sqrt{2}}\right)^2 +(x-y)^2= 1$$

$$\Rightarrow 3x^2 − 2xy + 3y^2 = 2$$

Answer 2

Using short-hand $s,c$ for $\sin \theta, \cos \theta$ $$x= s/ \sqrt 2 + c/2; \, y = s/ \sqrt 2 - c/2 $$

Remember as formulas the results/identities:

$$ ( x+y)^2 +( x+y)^2 = 2(x^2 +y^2);\, ( x+y)^2-( x+y)^2 = 4 x y ; $$

$$ 3 (x^2 +y^2) = 3 s^2 + 3 c^2/2$$

$$ -2 x y = c^2/2 - s^2 $$

Add, RHS

$$\rightarrow 2$$

It represents the equation of an ellipse originally without $xy$ term rotated by $45^0$ thereby bringing in the $xy$ term.