Ellipse Calculator - EMathHelp

Solution

The equation of an ellipse is $$$\frac{\left(x - h\right)^{2}}{a^{2}} + \frac{\left(y - k\right)^{2}}{b^{2}} = 1$$$, where $$$\left(h, k\right)$$$ is the center, $$$a$$$ and $$$b$$$ are the lengths of the semi-major and the semi-minor axes.

Our ellipse in this form is $$$\frac{\left(x - 0\right)^{2}}{9} + \frac{\left(y - 0\right)^{2}}{4} = 1$$$.

Thus, $$$h = 0$$$, $$$k = 0$$$, $$$a = 3$$$, $$$b = 2$$$.

The standard form is $$$\frac{x^{2}}{3^{2}} + \frac{y^{2}}{2^{2}} = 1$$$.

The vertex form is $$$\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$$$.

The general form is $$$4 x^{2} + 9 y^{2} - 36 = 0$$$.

The linear eccentricity (focal distance) is $$$c = \sqrt{a^{2} - b^{2}} = \sqrt{5}$$$.

The eccentricity is $$$e = \frac{c}{a} = \frac{\sqrt{5}}{3}$$$.

The first focus is $$$\left(h - c, k\right) = \left(- \sqrt{5}, 0\right)$$$.

The second focus is $$$\left(h + c, k\right) = \left(\sqrt{5}, 0\right)$$$.

The first vertex is $$$\left(h - a, k\right) = \left(-3, 0\right)$$$.

The second vertex is $$$\left(h + a, k\right) = \left(3, 0\right)$$$.

The first co-vertex is $$$\left(h, k - b\right) = \left(0, -2\right)$$$.

The second co-vertex is $$$\left(h, k + b\right) = \left(0, 2\right)$$$.

The length of the major axis is $$$2 a = 6$$$.

The length of the minor axis is $$$2 b = 4$$$.

The area is $$$\pi a b = 6 \pi$$$.

The circumference is $$$4 a E\left(\frac{\pi}{2}\middle| e^{2}\right) = 12 E\left(\frac{5}{9}\right)$$$.

The focal parameter is the distance between the focus and the directrix: $$$\frac{b^{2}}{c} = \frac{4 \sqrt{5}}{5}$$$.

The latera recta are the lines parallel to the minor axis that pass through the foci.

The first latus rectum is $$$x = - \sqrt{5}$$$.

The second latus rectum is $$$x = \sqrt{5}$$$.

The endpoints of the first latus rectum can be found by solving the system $$$\begin{cases} 4 x^{2} + 9 y^{2} - 36 = 0 \\ x = - \sqrt{5} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the first latus rectum are $$$\left(- \sqrt{5}, - \frac{4}{3}\right)$$$, $$$\left(- \sqrt{5}, \frac{4}{3}\right)$$$.

The endpoints of the second latus rectum can be found by solving the system $$$\begin{cases} 4 x^{2} + 9 y^{2} - 36 = 0 \\ x = \sqrt{5} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the second latus rectum are $$$\left(\sqrt{5}, - \frac{4}{3}\right)$$$, $$$\left(\sqrt{5}, \frac{4}{3}\right)$$$.

The length of the latera recta (focal width) is $$$\frac{2 b^{2}}{a} = \frac{8}{3}$$$.

The first directrix is $$$x = h - \frac{a^{2}}{c} = - \frac{9 \sqrt{5}}{5}$$$.

The second directrix is $$$x = h + \frac{a^{2}}{c} = \frac{9 \sqrt{5}}{5}$$$.

The x-intercepts can be found by setting $$$y = 0$$$ in the equation and solving for $$$x$$$ (for steps, see intercepts calculator).

x-intercepts: $$$\left(-3, 0\right)$$$, $$$\left(3, 0\right)$$$

The y-intercepts can be found by setting $$$x = 0$$$ in the equation and solving for $$$y$$$: (for steps, see intercepts calculator).

y-intercepts: $$$\left(0, -2\right)$$$, $$$\left(0, 2\right)$$$

The domain is $$$\left[h - a, h + a\right] = \left[-3, 3\right]$$$.

The range is $$$\left[k - b, k + b\right] = \left[-2, 2\right]$$$.