3

How do I show that the parametric equations

$$x(t) = \sin(t+a)$$

$$y(t) = \sin(t+b)$$

define an ellipse?

I tried graphing it and I'm certain it is a rotated ellipse. My first idea is to write it as

$$x(t) = \cos a \sin t + \sin a \cos t$$ $$y(t) = \cos b \sin t + \sin b \cos t$$

so the vector (x,y) is the vector (cos t, sin t) left multiplied by the matrix

$$\begin{pmatrix} \sin a & \cos a \\ \sin b & \cos b \end{pmatrix}$$

If I can show that this matrix is actually some enlargement followed by a rotation, I'm done, but I can't do this (I get contradictions).

Mhenni Benghorbal
  • 47,431
0998042
  • 95
  • Two what looked like minus signs went missing in the last edit. I have replaced them. If they should have disappeared, then edit them back out. – user1729 May 06 '13 at 12:18
  • What course is this? – Mhenni Benghorbal May 06 '13 at 14:02
  • If you really had $M = PD$ where $D$ is diagonal and $P$ is orthogonal, then that would force $M^tM = D^2$ which is again diagonal. But $M^tM$ is not diagonal (unless $a$ and $\pm b$ differ by $\pi/4$, or something similar), so you're getting contradictions. – ronno May 06 '13 at 14:29

3 Answers3

1

Here is how you advance

$$ x = \sin(t+a) \implies t=\sin^{-1}(x)-a . $$

Substituting in the other equation gives

$$ y = \sin( \sin^{-1}(x)+ b -a ) = \sin(\sin^{-1}(x))\cos(b-a)+\cos(\sin^{-1}(x))\sin(b-a) $$

$$ \implies y = x \cos(b-a) + \sqrt{1-x^2}\sin(b-a) $$

$$ \implies y-x \cos(b-a)=\sqrt{1-x^2}\sin(b-a) $$

$$ y^2+x^2-2\cos(b-a)\,xy-\sin^2(b-a)=0 $$

Now, Comparing with conic section general equation

$$ Ax^2 + Bxy + Cy^2 +Dx + Ey + F = 0\text{ with }A, B, C\text{ not all zero,}$$

we have

$$ A = 1,\quad B = -2\cos(b-a), \quad C = 1. $$

The equation represents an ellipse if $B^2-4AC<0$, that is

$$ 4\cos^2(b-a)-4= -4(1-\cos^2(b-a))= -4\sin^2(b-a)<0. $$

Note:

$$ \cos(\sin^{-1}(x))= \sqrt{ 1- \sin(\sin^{-1}(x))^2 }=\sqrt{1-x^2}. $$

Mhenni Benghorbal
  • 47,431
  • That's a somewhat awkward way to do it, going via $\sin^{-1}$. I recommend you edit it to do it a little more directly (rather than via inverse functions), by letting $s=t+a$ and then expressing $\sin(t+b)$ in terms of $\sin s$, etc. – Glen O May 06 '13 at 12:38
  • @GlenO: Why do you think it is an awkward way while it is simple and clear. – Mhenni Benghorbal May 06 '13 at 13:10
  • Because it requires the reader to understand how to handle terms like $\cos(\sin^{-1}(x))$ in general - something that isn't necessarily taught when a person is learning geometry. The basic trig rules, however, should be well-understood at that point. I recognise that it's personal opinion, but I felt I should say it anyway. – Glen O May 06 '13 at 13:13
  • 1
    @GlenO: It is pre-calculus stuff. Yet, I already put a note about it. I do not see this as a major problem. – Mhenni Benghorbal May 06 '13 at 13:20
  • Why is it okay to square both sides in the seventh line? I believe the equation in the sixth line is half an ellipse but when we square it, it becomes an ellipse. – 0998042 May 06 '13 at 14:09
  • @0998042 It comes down to sin^{-1} not being an inverse of sin on the whole circle. It's just a convention to take its range between -pi/2 and pi/2. If we were to be fully rigorous, we would write out two cases in the lines above; cos(sin^{-1}(x)) which is already there, as well as cos(\pi-sin^{-1}(x)). But after the squaring step, the two cases would merge into the same end result. – Matthew C Oct 15 '16 at 14:48
1

The parametrization represents an ellipse centered at the origin, albeit tilted with respect to the axes. (You can demonstrate by plotting a few for yourself.) The general form of this ellipse is

$$A x^2 + B x y + C y^2 = 1$$

The idea is to find the coefficients; this is done by expanding the sines and forming $x^2$, $y^2$, and $x y$. I leave the algebra to the reader; I get

$$\left (\begin{array} \\ \cos^2{a} & \cos{a} \cos{b} & \cos^2{b} \\ \cos{2 a} & \cos{(a+b)} & \cos{2 b} \\\sin{2 a} & \sin{(a+b)} & \sin{2 b} \end{array}\right ) \cdot \left ( \begin{array}\\A\\B\\C \end{array} \right ) = \left ( \begin{array}\\1\\0\\0 \end{array} \right )$$

One nontrivial point I should mention is that you get coefficients of $\sin^2{t}$, $\cos^2{t}$, and $\sin{t} \cos{t}$. To get an expression that can be set to $1$ on the RHS, I took $\cos^2{t} = 1-\sin^2{t}$ and that did the trick. You then get an expression independent of $t$ set to $1$, a coefficient of $\sin^2{t}$ set to zero, and a coefficient of $\sin{t} \cos{t}$, set to zero.

Inverting this matrix and performing the multiplication by the RHS, then rearranging the resulting equation for the ellipse, I get

$$x^2+y^2-2 \cos{(a-b)} x y = \sin^2{(a-b)}$$

as the sought ellipse. Note that this is indeed an ellipse because $\cos^2{(a-b)} < 1$.

I also note that one finds parametrization like this in discussions of elliptical polarization of light. Note that when $a-b = \pi/2$, the expression reduces to a circle; this is what physicists call circular polarization.

Ron Gordon
  • 138,521
0

The matrix approach will be nicer but here is a brute force way.

$x=\sin(t+a)=\sin t \cos a+\cos t \sin a$

$y=\sin(t+b)=\sin t \cos b+\cos t \sin b$

solve for $\sin t$ and $\cos t$

$y\cos a - x\cos b = \cos t (\sin b \cos a - \cos b \sin a) $

$y\sin a - x\sin b = \sin t (\sin a \cos b - \cos a \sin b) $

Now square the last two and add to get

$x^2+y^2-2xy \cos(a-b)= \sin^2(a-b)$

Now you need to figure angle of rotation, etc.

Maesumi
  • 3,702