Hyperbola Calculator - EMathHelp

Solution

The equation of a hyperbola 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 hyperbola in this form is $$$\frac{\left(x - 0\right)^{2}}{36} - \frac{\left(y - 0\right)^{2}}{9} = 1$$$.

Thus, $$$h = 0$$$, $$$k = 0$$$, $$$a = 6$$$, $$$b = 3$$$.

The standard form is $$$\frac{x^{2}}{6^{2}} - \frac{y^{2}}{3^{2}} = 1$$$.

The vertex form is $$$\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1$$$.

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

The linear eccentricity (focal distance) is $$$c = \sqrt{a^{2} + b^{2}} = 3 \sqrt{5}$$$.

The eccentricity is $$$e = \frac{c}{a} = \frac{\sqrt{5}}{2}$$$.

The first focus is $$$\left(h - c, k\right) = \left(- 3 \sqrt{5}, 0\right)$$$.

The second focus is $$$\left(h + c, k\right) = \left(3 \sqrt{5}, 0\right)$$$.

The first vertex is $$$\left(h - a, k\right) = \left(-6, 0\right)$$$.

The second vertex is $$$\left(h + a, k\right) = \left(6, 0\right)$$$.

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

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

The length of the major axis is $$$2 a = 12$$$.

The length of the minor axis is $$$2 b = 6$$$.

The focal parameter is the distance between the focus and the directrix: $$$\frac{b^{2}}{c} = \frac{3 \sqrt{5}}{5}$$$.

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

The first latus rectum is $$$x = - 3 \sqrt{5}$$$.

The second latus rectum is $$$x = 3 \sqrt{5}$$$.

The endpoints of the first latus rectum can be found by solving the system $$$\begin{cases} x^{2} - 4 y^{2} - 36 = 0 \\ x = - 3 \sqrt{5} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the first latus rectum are $$$\left(- 3 \sqrt{5}, - \frac{3}{2}\right)$$$, $$$\left(- 3 \sqrt{5}, \frac{3}{2}\right)$$$.

The endpoints of the second latus rectum can be found by solving the system $$$\begin{cases} x^{2} - 4 y^{2} - 36 = 0 \\ x = 3 \sqrt{5} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the second latus rectum are $$$\left(3 \sqrt{5}, - \frac{3}{2}\right)$$$, $$$\left(3 \sqrt{5}, \frac{3}{2}\right)$$$.

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

The first directrix is $$$x = h - \frac{a^{2}}{c} = - \frac{12 \sqrt{5}}{5}$$$.

The second directrix is $$$x = h + \frac{a^{2}}{c} = \frac{12 \sqrt{5}}{5}$$$.

The first asymptote is $$$y = - \frac{b}{a} \left(x - h\right) + k = - \frac{x}{2}$$$.

The second asymptote is $$$y = \frac{b}{a} \left(x - h\right) + k = \frac{x}{2}$$$.

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

x-intercepts: $$$\left(-6, 0\right)$$$, $$$\left(6, 0\right)$$$

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

Since there are no real solutions, there are no y-intercepts.