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Given the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$, find a trajectory $\sigma(t)$ which represent it.

So far, I have this:

The standard equation for an ellipse is:

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

So,

$4x^{2}+9y^{2}=36\}$

$=\left \langle \text{Arithmetic: Divide by}\ 36 \right \rangle$

$\frac{4x^{2}}{36}+\frac{9y^{2}}{36}=\frac{36}{36}$

$=\left \langle \text{Arithmetic} \right \rangle$

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$

I don't know how to continue developing the solution, but I think the trajectory could be represented as $(a\cos(t),b\sin(t))$

Some ideas, suggestions to solve this?

InfZero
  • 875

2 Answers2

2

You are correct, an ellipse of the form $$ \frac{x^2}{a^2}+\frac{y^2}{b^2} =1 $$ has parametric equations \begin{cases} x=a\cos t\\ y = b\sin t \end{cases} with $t\in [0,2\pi]$. So in your case, you can use \begin{cases} x={3}\cos t\\ y = {2}\sin t \end{cases}

Kuifje
  • 9,584
1

For a circle $$x^2+y^2 = r^2$$

we can try

$$x=r\cos(t)$$ $$y=r\sin(t)$$

because

$$x^2+y^2 = r^2\cos^2(t)+r^2\sin^2(t) = r^2$$

So for an ellipse

$$a^2x^2+b^2y^2 = c^2$$

we can try

$$x=\frac{c}{a}\cos(t)$$ $$y=\frac{c}{b}\sin(t)$$

In our case we have

$$2^2x^2+3^2y^2 = 6^2$$

so let us try try

$$x=\frac{6}{2}\cos(t)$$ $$y=\frac{6}{3}\sin(t)$$

Trying:

$$2^2x^2+3^2y^2 = 2^2(\frac{6}{2}\cos(t))^2+3^2(\frac{6}{3}\sin(t))^2$$

$$ = 6^2\cos^2(t) + 6^2\sin^2(t) = 6^2 = 36$$

BCLC
  • 13,459