Why can we use $\cos^2t+\sin^2t=1$ to eliminate $t$ from $x=5\cos t$, $y=2\sin t$?

Question

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.

Answer 1

There are two questions here.

Question 1: is every point that can be written as $x = 5 \cos t$, $y = 2 \sin t$ on the ellipse $x^2 / 25 + y^2 / 4 = 1$?

The answer to this question is "yes". This is clearly true by the trigonometric identity.

Question 2: can every point $(x, y)$ such that $x^2 / 25 + y^2 / 4 = 1$ be written as $x = 5 \cos t$ and $y = 2 \sin t$?

The answer to this question is also "yes", but it's less obvious (and you haven't yet proved it).

So you currently know that every point on your parametrized curve is on this ellipse, but you haven't yet proved that every point on the ellipse is on your curve.

Answer 2

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.

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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.

Answer 3

A more expedite method to eliminate $t$ is to set a value of $x$, solve for $t$, and plug in $y$.

$$x=5\cos t\to t=2k\pi\pm\arccos\frac x5\to y=2\sin\left(2k\pi\pm\arccos\frac x5\right)=\pm\sin\left(\arccos\frac x5\right).$$

We can rework this by squaring, and we establish

$$\frac{y^2}{2^2}=\sin^2\left(\arccos\frac x5\right)=1-\sin^2\left(\arcsin\frac x5\right)=1-\frac{x^2}{5^2}.$$

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An additional question is to kwow the ranges of $x$ and $y$ values that are reached, but as the range of $t$ was not specified, we can't say more.