Why can we use $\cos^2t+\sin^2t=1$ to eliminate $t$ from $x=5\cos t$, $y=2\sin t$? Given the parametric equations $x=5 \cos t$ and $y=2 \sin t$, I want to eliminate the parameter.
So, how can we eliminate the parameter here? Supposedly, all we need to do according to the solution is to recall this trig identity:
$$
\cos ^{2} t+\sin ^{2} t=1
$$
Then from the parametric equations we get,
$$
\cos t=\frac{x}{5} \quad \sin t=\frac{y}{2}
$$
Then, using the trig identity from above and these equations we get,
$$
1=\cos ^{2} t+\sin ^{2} t=\left(\frac{x}{5}\right)^{2}+\left(\frac{y}{2}\right)^{2}=\frac{x^{2}}{25}+\frac{y^{2}}{4}
$$ and should thus conclude that we haven an ellipse.

However, I don't understand why this works. We just took some equation (trig identity) and plugged something in - how do we know that this is equal to our parametric equations? I mean we could have taken any other formula and plug in values and make a completely different conclusion - it seems very arbitrary to me.


This is because $\left(\frac{x}{5}\right)^{2} + \left(\frac{y}{2}\right)^{2}$ matches the form of $\cos^{2}t + \sin^{2}t$. But before this, let's understand what $\cos^{2}t + \sin^{2}t = 1$ mean.

We know that the center-radius (standard) form of a circle of radius $r$ with its center at the origin has the equation $$x^{2} + y^{2} = r^{2}.$$
Dividing both sides by $r^{2}$,
$$\begin{align*}\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} &= 1 \\ \left(\frac{x}{r}\right)^{2} + \left(\frac{y}{r}\right)^{2} &= 1.\end{align*}$$
But we know that $\cos t =\frac{x}{r}$ and $\sin t = \frac{y}{r}$ where $t$ is the angle from the positive $x$-axis. By substitution, we get the equation $$\cos^{2}t + \sin^{2}t = 1 \tag{1}.$$
We will now go to ellipses. Circles as special cases of ellipses where both $x$ and $y$ are scaled by a factor of $r$. However, ellipses that are not circles have different scaling factors for $x$ and $y$. If the scaling factor for $x$ and $y$ are $a$ and $b$, the equation will be $$\left(\frac{x}{a}\right)^{2} + \left(\frac{y}{b}\right)^{2} = 1 \tag{2}.$$
As $(1)$ is similar to $(2)$, we can equate the terms to each other.
\begin{align*}\left(\frac{x}{a}\right)^{2} &= \cos^{2}t &\qquad \left(\frac{y}{b}\right)^{2} &= \sin^{2}t \\ \frac{x}{a} &= \cos t &\qquad \frac{y}{b} &= \sin t \\ x &= a\cos t &\qquad y &= b \sin t.\end{align*}
This is why it works. I can't think of anything aside from this.
