Question

write an equation for an ellipse centered at the origin, which has foci at (+-3,0) and co vertices at (0+-4)

Answer 1

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:

$\frac{x^{2}}{7} + \frac{y_{2}}{16} = 1$

Step-by-step explanation:

An ellipse center at origin is modelled after the following expression:

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

Where:

$a$, $b$ - 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:

$2\cdot b = \sqrt{(0-0)^{2}+ [4 - (-4)]^{2}}$

$2\cdot b = \pm 8$

$b = \pm 4$

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:

$2\cdot c = \sqrt{[3 - (-3)]^{2}+ (0-0)^{2}}$

$2\cdot c = \pm 6$

$c = \pm 3$

Now, the length of the minor semi-axis is given by the following Pythagorean identity:

$a = \sqrt{b^{2}-c^{2}}$

$a = \sqrt{4^{2}-3^{2}}$

$a = \pm \sqrt{7}$

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:

$\frac{x^{2}}{7} + \frac{y_{2}}{16} = 1$