Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$

Question

Eliminate $\theta$ in following equations
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$

I am trying to solve this problem but still I am unable to get the perfect answer
I added both the equations but it transformed it to
$2 \cos(\theta+(\alpha + \beta)/2)$

Answer 1

\begin{align} \frac{x}{a}+\frac{y}{b} &= \cos (\theta-\alpha)+\cos (\theta-\beta) \\ &=2\cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta-2\theta}{2} \\ \frac{x}{a}-\frac{y}{b} &= \cos (\theta-\alpha)-\cos (\theta-\beta) \\ &=-2\sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta-2\theta}{2} \\ 1 &= \left( \frac{\frac{x}{a}+\frac{y}{b}}{2\cos \frac{\alpha-\beta}{2}} \right)^2+ \left( -\frac{\frac{x}{a}-\frac{y}{b}}{2\sin \frac{\alpha-\beta}{2}} \right)^2 \\ \sin^2 (\alpha-\beta) &= \frac{x^2}{a^2}-\frac{2xy\cos (\alpha-\beta)}{ab}+\frac{y^2}{b^2} \end{align}

• The curve is known as Lissajous figure (with same frequencies).

• The area bounded by the ellipse is $A=\pi ab\sin (\beta-\alpha)$.

• $A>0$ gives anti-clockwise trace whereas $A<0$ for clockwise.

• When the curve degenerates to a line segment, $A=0$