Eliminating parameter from 2 parametric equations

Question

I am given two parametric equations as-

$ x= g\sin^2(t) - f\sin(t)\cos(t) $

$ y = f\cos^2(t) - g\sin(t)\cos(t) $

and I have to eliminate the parameter $t$.

I tried taking $\sin(t)$ common from the first equation and $\cos(t)$ from the second, and divided the two to get $ x/y = -\tan(t) $ but I couldn't proceed from there.

Answer 1

Hint: using $\sin(2t)=2 \sin(t)\cos(t)$ and $\cos(2t)=2 \cos^2(t)-1=1-2\sin^2(2t)\,$: $$ f x + g y = fg - \frac{f^2+g^2}{2} \sin(2t) \\ g x - f y = \frac{g^2-f^2}{2}-\frac{f^2+g^2}{2}\cos(2t) $$

Then use $\sin^2(2t)+\cos^2(2t)=1$ to eliminate $t$ between the two equations.

Answer 2

\begin{align} \cos^2 t & = \frac 1 2 + \frac 1 2 \cos(2t) \\[10pt] \sin^2 t & = \frac 1 2 - \frac 1 2 \cos(2t) \\[10pt] \sin(t) \cos(t) & = \frac 1 2 \sin(2t) \\[10pt] x & = g\sin^2 t - f\sin(t) \cos(t) = g\left( \frac 1 2 - \frac 1 2\cos(2t) \right) - f \cdot \frac 1 2 \sin(2t) \\[10pt] y & = f \cos^2 t - g \sin(t)\cos(t) = f\left( \frac 1 2 + \frac 1 2 \cos(2t) \right) - g\cdot \frac 1 2 \sin(2t) \\[10pt] fx - gy & = -fg\cos(2t) + \frac{g^2- f^2} 2 \sin(2t) \\[10pt] gx - fy & = \frac{g^2-f^2} 2 - \frac {g^2+f^2} 2 \cos(2t) \\[10pt] \sin(2t) & = \cdots \tag A \\[10pt] \cos(2t) & = \cdots \tag B \\[10pt] \text{Therefore } & \left( \cdots \right)^2 + \left( \cdots \right)^2 = 1. \text{ (Here insert the expressions in lines (A) and (B).)} \end{align}