Solving the following parametric equation

Question

Solve the following parametric equation:

$$\frac{-(3\cos t-x)}{2\sin t-y}=-\frac{2\cos t}{3\sin t}$$

So I need to find the parametric equation of the thing in terms of $t$. But I am confused as to how to do it since I have two variables but only one equation.

If anyone could help, it would mean a lot! Thank you :)

Answer 1

Rewriting into:

$$y-2\sin t=\frac{3\sin t}{2\cos t} (x-3\cos t) \quad \cdots \cdots \: (*)$$

which is an equation of normal of the ellipse

$$\frac{x^2}{3^2}+\frac{y^2}{2^2}=1$$

The envelope of $(*)$ is the centre of curvature of the ellipse (or ellipse evolute):

\begin{align*} (x,y) &= \left( \frac{3^2-2^2}{3} \cos^{3} t, \frac{2^2-3^2}{2} \sin^{3} t \right) \\ &= \left( \frac{5}{3} \cos^{3} t, -\frac{5}{2} \sin^{3} t \right) \end{align*}

Equation of normal for $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(a\cos t,b\sin t)$ is

$$ax\sec t-by\csc t=a^2-b^2$$