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This is not homework, it's something that I didn't understood:

Given the function $f:\mathbb{R}\to \mathbb{R^2}$ such that $$f(t)=\left(\frac{2c^2\cos(t)}{1+\sin^2(t)},\frac{2c^2\sin(t)\cos(t)}{1+\sin^2(t)}\right)$$ where $c$ is a constant.

  1. How do we know this is a parametrization of the lemniscate?

  2. What happens with the derivative at (0,0)

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Recognizing what is a parametrized curve is, in many cases, difficult (at least to me). Plotting the function for any given value of $c$ would show what it is (as long as you already know the shape of a lemniscate).

If you look here, you will notice that it is effectively a lemniscate of Bernoulli.

Considering the derivatives, using trigonometric identities to simplify them, you should find $$\frac{dx}{dt}=-\frac{c^2 \sin (t) (\cos (2 t)+5)}{\left(1+\sin ^2(t)\right)^2}$$ $$\frac{dy}{dt}=\frac{c^2 (3 \cos (2 t)-1)}{\left(1+\sin ^2(t)\right)^2}$$ which make $$\frac{dy}{dx}=\frac {\frac{dy}{dt}} {\frac{dx}{dt}}=-\frac{(3 \cos (2 t)-1) \csc (t)}{\cos (2 t)+5}$$ In the range $0\leq t \leq 2\pi$, we have $x=y=0$ if $t=\frac \pi 2$ and $t=\frac {3\pi} 2$. For each of these values compute $\frac{dy}{dx}$.

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The symbolism $(x,y)$ represents the coordinates of a point. Thus : $$f(t)=\left(\frac{2c^2\cos(t)}{1+\sin^2(t)},\frac{2c^2\sin(t)\cos(t)}{1+\sin^2(t)}\right)=\left(x(t),y(t)\right)$$ This means that the position of point $(x,y)$ depends on the parameter $t$ : this is the definition of the parametric equation of a curve, where : $$\begin{cases} x(t)=\frac{2c^2\cos(t)}{1+\sin^2(t)} \\ y(t)=\frac{2c^2\sin(t)\cos(t)}{1+\sin^2(t)} \end{cases}$$ Of course, it is necessary to know what is a lemniscate if we want to recognize a lemniscate. One can find this in the mathematical handbooks together with the equation of lemniscate.

For example in http://mathworld.wolfram.com/Lemniscate.html , comparing the equations (9), (10) to the above parametric equation, with $a=2c^2$ allows to conclude that the curve is a lemniscate.

At particular values of $t=t_n=\frac{\pi}{2}+n\pi \quad\to\quad \cos(t_n)=0\quad$ hence $f(t_n)=(0,0)$

This means that the curve crosses the point origin $(0,0)$.

For more description of the curve close to the origin, we have to compute the derivatives $x'(t_n)$ and $y'(t_n)$. We find : $$\begin{cases} x'(t_n)=-2c^2 \text{ if }n\text{ even , and}\quad =2c^2 \text{ if }n\text{ odd} \\ y'(t_n)=-2c^2 \end{cases}$$ The slope of the curve $\frac{dy}{dx}=\frac{y'}{x'}$ is $1$ for $n$ even and $-1$ for $n$ odd. So, the curve crosses the origin two times respectively with slope $1$ and $-1$.

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