Respuesta :
The given equation of the ellipse is:
[tex]x^2+4y^2\text{ = 4}[/tex]To find the x intercept, let y = 0
[tex]\begin{gathered} x^2+4(0)^2\text{ = 4} \\ x\text{ = }\sqrt[]{4} \\ x\text{ = }\pm2 \\ x-\text{intercept = (}\pm2,\text{ 0)} \end{gathered}[/tex]For the y-intercept, let x = 0
[tex]\begin{gathered} 0^2+4y^2=4 \\ y^2\text{ = }\frac{4}{4} \\ y\text{ = }\sqrt[]{1} \\ y\text{ = }\pm1 \\ y-\text{intercept = (0, }\pm1) \end{gathered}[/tex]For an ellipse of the form:
[tex]\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \text{The x-domain is from -a and a} \\ \text{The y-domain is from -b and b} \end{gathered}[/tex]The given equation is:
[tex]\begin{gathered} x^2+4y^2=\text{ 4} \\ \text{Divide through by 4} \\ \frac{x^2}{4}+\frac{y^2}{1}=\frac{4}{4} \\ \frac{x^2}{4}+\frac{y^2}{1}=\text{ 1} \end{gathered}[/tex][tex]\begin{gathered} a^2\text{ = 4} \\ a\text{ = }\sqrt[]{4} \\ a\text{ = -2, 2} \\ b^2\text{ = 1} \\ b\text{ = }\sqrt[]{1} \\ b\text{ = -1, 1} \end{gathered}[/tex]The x-domain = (-2, 2)
The y-domain = (-1, 1)
Symmetry test:
For the equation to be symmetriacal to the x-axis, y → -y
[tex]\begin{gathered} x^2+4y^2\text{ = 4} \\ y^2\text{ = 1 - }\frac{x^2}{4}\ldots\ldots\text{.}(1) \\ \text{Let y = -y} \\ (-y)^2\text{ = 1 - }\frac{x^2}{4} \\ y^2\text{ = 1 - }\frac{x^2}{4}\ldots\ldots\text{.}\mathrm{}(2) \\ \end{gathered}[/tex]Because (1) matches with (2), the equation is symmetrical to the x-axis
For the equation to be symmetrical to the y-axis, x → -x
[tex]\begin{gathered} x^2+4y^2\text{ = 4}\ldots..(1) \\ \text{Let x = -x} \\ (-x)^2+4y^2\text{ = 4} \\ x^2+4y^2\text{ = 4}\ldots.(2) \end{gathered}[/tex]Since (1) matches with (2), the equation is symmetrical to the y-axis
For the equation to be symmetrical to the origin, x→-x, y→-y
[tex]\begin{gathered} (-x)^2+4(-y)^2\text{ = 4} \\ x^2+4y^2\text{ = 4} \end{gathered}[/tex]The equation is symmetrical to the origin
Therefore, according to the symmetry test, the equation given is symmetrical to the x-axis, y-axis, and the origin
