# Finding cartesian equation for trigonometric parametric forms

I'm trying to find the cartesian equation for these parameteric forms:

$$x = sin\theta + 2 cos \theta \\ y = 2 sin\theta + cos\theta$$

I tried:

\begin{align} x^2 & = sin^2\theta + 4cos^2\theta \\ & = 1 - cos^2\theta + 4cos^2\theta \\ & = 1 + 3cos^2\theta \\ \\ y^2 & = 4sin^2\theta+cos^2\theta\\ & = 4(1 - cos^2\theta) + cos^2\theta \\ & = 4 - 3cos^2\theta \\ \\ \therefore & \space4 - y^2 = x^2 - 1\\ \space & x^2 + y^2 = 5 \end{align}

Which differs from the given answer of $5x^2 + 5y^2 - 8xy = 9$. Where am I going wrong?

Note that $$x^2\not=\sin^2\theta+4\cos^2\theta$$ and that $$x^2=\sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta$$ because $$(a+b)^2=a^2+b^2+\color{red}{2ab}.$$
Since we can get $$\sin\theta=\frac{-x+2y}{3},\ \ \ \cos\theta=\frac{2x-y}{3},$$ then use $$\sin^2\theta+\cos^2\theta=1.$$
• Got it. Finding $sin\theta$ and $cos\theta$ first made it so much easier :) – hohner Sep 28 '14 at 10:23
In general, $(a+b)^2\ne a^2+b^2$ unless $ab=0$
Solve for $\sin\theta,\cos\theta$ in terms of $x,y$
Then use $\sin^2\theta+\cos^2\theta=1$ to eliminate $\theta$