Writing a parametric equation in $y=f(x)$ form

Question

I have two equations in $x$ and $y$ in terms of some parameter $t$ as under $$ x(t)=-a.sin(at)-a\ \text{and}\\ y(t)=a.cos(at) .$$ Is it possible to eliminate t and find the relation between $x$ and $y$ in $y=f(x)$ or $f(x,y)=0$?Are there some general methods to achieve that end.Thank you in adavance for your responces and comments.

Answer 1

$$ x(t)=(-a\sin(at)-a) \\ \text{and} \\ y(t)=a\cos(at). $$ $$ \left\{ \begin{array}{l} x(t)=-a\sin(at)-a,\\ y(t)=a\cos(at).\\ \end{array} \right. \iff \left\{ \begin{array}{l} -\frac{x(t)}{a}-1=\sin(at),\\ \frac{y(t)}{a}=\cos(at).\\ \end{array} \right. \iff \left\{ \begin{array}{l} t=\frac{\arcsin\left(-\frac{x(t)}{a}-1\right)}{a},\\ \frac{y(t)}{a}=\cos(at).\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} t=\frac{\arcsin\left(-\frac{x(t)}{a}-1\right)}{a},\\ \frac{y(t)}{a}=\cos({\arcsin\left(-\frac{x(t)}{a}-1\right)}).\\ \end{array} \right. \iff \left\{ \begin{array}{l} t=\frac{\arcsin\left(-\frac{x(t)}{a}-1\right)}{a},\\ \frac{y(t)}{a}=\left(1-\left(-\frac{x(t)}{a}-1\right)^{2}\right)^{0.5}.\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} t=\frac{\arcsin\left(-\frac{x(t)}{a}-1\right)}{a},\\ {y}=a\left(\frac{x^{2}}{a^{2}}-\frac{2x}{a}\right)^{0.5}.\\ \end{array} \right. $$ So we get that $$ \bbox[lightgreen] { {y}=a\left(\frac{x^{2}}{a^{2}}-\frac{2x}{a}\right)^{0.5}. } $$ Good luck!

Answer 2

Due to the transcendental nature of equations $a$ cannot be eliminated resulting into elementary functions. We use numerical methods to find such solutions. E.g., Newton-Raphson, Regula Falsi etc.

Using identity $ \sin ^2 at + \cos^2 at =1 $ we obtain the eliminant Circle without a need to carry out such an elimination.

$$ \frac{(x+a)^2}{a^2}+ \frac{y^2}{a^2}=1$$