Parametric equation for the ellipse

Question

Let $x = A\sin(at+\theta)$ and $y = A\sin(at)$ . Prove that this parametric equation forms an ellipse except when $\theta = 0 $ and $\pi$ .

My try : I expanded the $\sin(at + \theta)$ and looked for a useful relation between $x$ and $y$ but didn't work . I think it should be a rotated ellipse and the rotation matrix is involved .

Answer 1

Let $t=u-\dfrac\theta{2a}$. Then $x=A\sin\left(au+\dfrac{\theta}{2}\right)$ and $y=A\sin\left(au-\dfrac{\theta}{2}\right)$.

$x+y=2A\sin au\cos\dfrac\theta2$ and $x-y=2A\cos au\sin\dfrac\theta2$.

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

It is an ellipse with axes making $45^\circ$ with the coordinate axes.

Answer 2

$x=A\sin at\cos\theta+B\cos at\sin\theta=y\cos\theta+B\sin\theta\cos at$

$\implies\cos at=?$

Now use $\cos^2at+\sin^2at=1$ to eliminate $t$

If $\theta=n\pi,$

$x=y+(-1)^n$ which represents straight line (s)

Answer 3

Write $x=A\sin (at) \cos\theta+A\cos(at)\sin\theta$

Also note that $A\sin(at) = y$, so $A\cos(at)=\sqrt{A^2-y^2}$. So we have

$$x = y\cos\theta+\sin\theta\sqrt{A^2-y^2}$$ $$x-y\cos\theta=\sin\theta\sqrt{A^2-y^2}$$ $$x^2+y^2\cos^2\theta-2xy\cos\theta=A^2\sin^2\theta-y^2\sin^2\theta$$ $$x^2+y^2-2xy\cos\theta=A^2\sin^2\theta$$

Rotating this conic at an angle $\displaystyle{\phi = \frac12\arctan\left(\frac{-2\cos\theta}{1-1}\right)} = 45^{o}$ (so as to remove the term with $xy$), with the transformation equations $$x = X\cos45^{o}-Y\sin45^{o}=\frac{X-Y}{\sqrt2},$$ $$Y=X\sin45^{o}+Y\cos45^{o}=\frac{X+Y}{\sqrt2}$$ We have $$(1-\cos\theta)X^2+(1+\cos\theta)Y^2=A^2\sin^2\theta=A^2\left(1-\cos^2\theta\right)$$ $$\frac{X^2}{1+\cos\theta}+\frac{Y^2}{1-\cos\theta}=A^2$$ wich is the equation of an ellipse except for $\theta = 0$ or $\theta =\pi$, which make the denominators of $X^2$ and $Y^2$ zero, respectively; resulting into an undefined situation.