Answer:
The equation for an ellipse centered at the origin with foci at (-3, 0) and (+3, 0) and co-vertices at (0, -4) and (0, +4) is:
[tex]\frac{x^{2}}{7} + \frac{y_{2}}{16} = 1[/tex]
Step-by-step explanation:
An ellipse center at origin is modelled after the following expression:
[tex]\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1[/tex]
Where:
[tex]a[/tex], [tex]b[/tex] - Major and minor semi-axes, dimensionless.
The location of the two co-vertices are (0, - 4) and (0, + 4). The distance of the major semi-axis is found by means of the Pythagorean Theorem:
[tex]2\cdot b = \sqrt{(0-0)^{2}+ [4 - (-4)]^{2}}[/tex]
[tex]2\cdot b = \pm 8[/tex]
[tex]b = \pm 4[/tex]
The length of the major semi-axes can be calculated by knowing the distance between center and any focus (c) and the major semi-axis. First, the distance between center and any focus is determined by means of the Pythagorean Theorem:
[tex]2\cdot c = \sqrt{[3 - (-3)]^{2}+ (0-0)^{2}}[/tex]
[tex]2\cdot c = \pm 6[/tex]
[tex]c = \pm 3[/tex]
Now, the length of the minor semi-axis is given by the following Pythagorean identity:
[tex]a = \sqrt{b^{2}-c^{2}}[/tex]
[tex]a = \sqrt{4^{2}-3^{2}}[/tex]
[tex]a = \pm \sqrt{7}[/tex]
The equation for an ellipse centered at the origin with foci at (-3, 0) and (+3, 0) and co-vertices at (0, -4) and (0, +4) is:
[tex]\frac{x^{2}}{7} + \frac{y_{2}}{16} = 1[/tex]
