Question

Give an example of a parametrization of the ellipse

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

in${\mathbf{ℝ}}^{\mathbf{2}}$ . See Example .

Short answer

$\left[\begin{array}{c}\cos \left(t\right)\\ 2\sin \left(t\right)\end{array}\right]$ is a parametrization of the ellipse$x^2+\frac{y^2}{4}=1$ in${\mathrm{ℝ}}^2$ .

Step-by-step solution

Consider the parameters.

The image of a function consists of all the values the function takes in its target space. If f is a function from X to Y, then

$$\begin{gathered} image\left(f\right)=\left\{f\left(x:xinX\right)\right\} \\ =\left\{binY:b=f\left(x\right),forsomexinX\right\} \end{gathered}$$

The standard form of an ellipse is,$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The parametrization of this ellipse is,

$$\begin{gathered} x=\left(a\right)\cos \left(t\right),y=\left(b\right)\sin \left(t\right) \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\left(a\right)\cos \left(t\right)\\ \left(b\right)\sin \left(t\right)\end{array}\right] \end{gathered}$$

The ellipse equation is, $\frac{x^2}{1}+\frac{y^2}{4}=1$

The parametrization of this ellipse is,

$$\begin{gathered} x=\left(1\right)\cos \left(t\right),Y=\left(2\right)\sin \left(t\right) \\ \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\left(1\right)\cos \left(t\right)\\ \left(2\right)\sin \left(t\right)\end{array}\right] \end{gathered}$$

Final answer.

$\left[\begin{array}{c}\cos \left(t\right)\\ 2\sin \left(t\right)\end{array}\right]$ is a parametrization of the ellipse$x^2+\frac{y^2}{4}=1$ in${\mathrm{ℝ}}^2$ .