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As in the title, if $\textbf{a}$ and $\textbf{b}$ are vectors in the $x-y$ plane, then the parametrised form (with parameter $t$) $\textbf{a} \cos \omega t + \textbf{b} \cos \omega t$ traces out an ellipse.

I wrote $\textbf{a} = a_x \textbf{i} + a_y \textbf{j}$ and similarly for $\textbf{b}$ and then we have $$x = a_x \cos \omega t + b_y \sin \omega t$$ and $$y = a_y \cos \omega t + b_y \sin \omega t$$

I write $\cos$ and $\sin$ as $c$ and $s$, respectively. I have to show that this can be written in the form $$Ax^2 + 2Bxy + Cy^2 = E$$ for some constants $A, B, C, E$.

I expand this out so we have $$ Ax^2 = A(a_x^2 c^2 + b_x^2s^2 + 2a_xb_x sc)$$ $$ Cy^2 = C(a_y^2 c^2 + b_y^2 c^2 + 2a_y b_y sc)$$ and $$ 2Bxy = 2B(a_x a_y c^2 + b_x b_y s^2 + a_x b_y sc + a_y b_x sc)$$ Here $E$ is not relevant. I have to eliminate the term with $sc$ and also use the identity $s^2 +c^2 =1$ the left hand side to equal a constant. The coefficient of $s^2$ has to equal that of $c^2$. So it looks like I have two equations $$ Aa_xb_x + Ca_y b_y + Ba_xb_y + Ba_yb_x = 0 $$ and $$Aa_x^2 +Ca_y^2 +2Ba_xa_y = Ab_x^2 + Cb_y^2 + 2Bb_xb_y$$

To solve for $A,B,C$ I need another equation and I'm not sure what that is.

user110503
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  • Thanks. I feel bit silly of me not to have done this. – user110503 Sep 04 '16 at 14:40
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    (Reposting comment to get rid of a typo.) Using the shorthand $$\delta=a_xb_y-a_yb_x$$ for the relevant Cramér determinant, note that $$b_yx-b_xy=\delta\cos(\omega t)\qquad a_yx-a_xy=-\delta\sin(\omega t)$$ hence, based on the fact that $\cos^2(\omega t)+\sin^2(\omega t)=1$ for every $t$, the equation you are after is $$(b_yx-b_xy)^2+(a_yx-a_xy)^2=\delta^2$$ which is indeed the equation of an ellipse except when $\delta=0$ or $a_x=b_x=0$ or $a_y=b_y=0$ (but then the set is not an ellipse anyway). In your notations, $$A=a_y^2+b_y^2\quad B=-b_xb_y-a_xa_y\quad C=a_x^2+b_x^2\quad E=\delta^2$$ – Did Sep 04 '16 at 14:41

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Consider the map $$(x, y) \mapsto x\mathbf a + y \mathbf b.$$ This transforms the unit circle into the set you're interested in.

N.B.: if $\mathbf a$ and $\mathbf b$ are linearly dependent, then the claim is false, unless you allow "a line segment" or "a single point" as special cases of an ellipse. So I'm going to assume they're independent.

Assuming that the two (bold) vectors are linearly independent, the image $E$ of the circle under such a transformation is always an ellipse, for if $S$ is the matrix representing the inverse of this transformation, we have that for any $\mathbf e \in E$, $$ \| S \mathbf e \| = 1 $$ which can be rewritten $$ \mathbf e^t S^t S \mathbf e = 1.$$

Now $S^t S$ is symmetric and positive definite, and hence diagonalizable with positive eignevalues, so $S^tS = U^t D U$ where $U$ is orthogonal and $D$ is diagonal with positive entries, so $S^t S = U^t H^t H U$ where $H$ is a diagonal matrix whose entries are the square roots of the diagonal entries of $D$, i.e. $H^2 = H^t H = D$. This gives us

$$ \mathbf e^t S^t S \mathbf e = 1\\ \mathbf e^t U^t H^t H U \mathbf e = 1. $$ Changing coordinates by $U$,i.e., $f = Ue$, you get $$ \mathbf f^t H^t H \mathbf f = 1. $$ which can be unravelled into the equation of an axis-aligned ellipse in the $U$-coordinate system.

Short form: Apply Sylvester's Law of Inertia (a deep theorem from linear algebra) to avoid writing a whole lot of algebraic stuff.

John Hughes
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  • Thanks for this. I have to say I have forgotten some things in linear algebra so I don't fully understand everything here.

    That being said would my approach work? Is there a way to get the last equation?

     – user110503 Sep 04 '16 at 14:27
  • See @Did's comment on your question. – John Hughes Sep 04 '16 at 14:47