A curve has parametric equations
$x=a \sin(t)+b \cos(t)$
$y=a \cos(t)-b \sin(t)$
How do I eliminate t to find the Cartesian equation here?
I've tried different weird approaches, i.e. squaring everything and trying to use a trig identity or rearranging and substituing one part of the equation into the other but I'm not getting anywhere and can't eliminate t. I'm not really good when it comes to parametric equations.
$x = a\sin t + b\cos t$
$y = a\cos t - b \sin t$
$x^2 + y^2 = (a\sin t + b\cos t)^2 + (a\cos t - b \sin t)^2$
$x^2 + y^2 = a^2(\sin^2 t + \cos^2t) + b^2(\sin^2 t + \cos^2t) + 2ab\cos t\sin t - 2ab\cos t \sin t$
$$x^2 + y^2 = a^2 + b^2$$
Think of $\sin t$ and $\cos t$ as variables, solve the system, then square and add them together. From the first equation, $$\sin t=\frac 1a(x-b\cos t)$$ Plug this into the second one: $$y=a\cos t-\frac ba(x-b\cos t)$$ After rearranging the terms:$$ya-bx=(a^2+b^2)\cos t$$ Can you take it from here?
Hint: $$x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin(t)+\frac{b}{\sqrt{a^2+b^2}}\cos(t)\right)$$ so $$\cos(\phi)=\frac{a^2}{\sqrt{a^2+b^2}}$$ and $$\sin(\phi)=\frac{b}{\sqrt{a^2+b^2}}$$
Hint:
The key here is to utilize $$\cos^2t+\sin^2t=1\ \ \ \ (1)$$ to eliminate $t$
Solve the simultaneous equations for $\cos t,\sin t$ and use $(1)$
Observe that $$bx+ay = (a^2+b^2)\cos t \\ ax-by = (a^2+b^2)\sin t.$$ Now you can use the identity $\cos^2t+\sin^2t=1$ by squaring both equations and adding them: $$(a^2+b^2)(x^2+y^2) = (a^2+b^2)^2,$$ which reduces to $x^2+y^2=a^2+b^2$ if $a^2+b^2\ne0$.
