Properties of the hyperbola $$$- 25 x^{2} + 4 y^{2} = 100$$$

The calculator will find the properties of the hyperbola $$$- 25 x^{2} + 4 y^{2} = 100$$$, with steps shown.

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Find the center, foci, vertices, co-vertices, major axis length, semi-major axis length, minor axis length, semi-minor axis length, latera recta, length of the latera recta (focal width), focal parameter, eccentricity, linear eccentricity (focal distance), directrices, asymptotes, x-intercepts, y-intercepts, domain, and range of the hyperbola $$$- 25 x^{2} + 4 y^{2} = 100$$$.

Solution

The equation of a hyperbola is $$$\frac{\left(y - k\right)^{2}}{b^{2}} - \frac{\left(x - h\right)^{2}}{a^{2}} = 1$$$, 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 $$$\frac{\left(y - 0\right)^{2}}{25} - \frac{\left(x - 0\right)^{2}}{4} = 1$$$.

Thus, $$$h = 0$$$, $$$k = 0$$$, $$$a = 2$$$, $$$b = 5$$$.

The standard form is $$$\frac{y^{2}}{5^{2}} - \frac{x^{2}}{2^{2}} = 1$$$.

The vertex form is $$$\frac{y^{2}}{25} - \frac{x^{2}}{4} = 1$$$.

The general form is $$$25 x^{2} - 4 y^{2} + 100 = 0$$$.

The linear eccentricity (focal distance) is $$$c = \sqrt{b^{2} + a^{2}} = \sqrt{29}$$$.

The eccentricity is $$$e = \frac{c}{b} = \frac{\sqrt{29}}{5}$$$.

The first focus is $$$\left(h, k - c\right) = \left(0, - \sqrt{29}\right)$$$.

The second focus is $$$\left(h, k + c\right) = \left(0, \sqrt{29}\right)$$$.

The first vertex is $$$\left(h, k - b\right) = \left(0, -5\right)$$$.

The second vertex is $$$\left(h, k + b\right) = \left(0, 5\right)$$$.

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

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

The length of the major axis is $$$2 b = 10$$$.

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

The focal parameter is the distance between the focus and the directrix: $$$\frac{a^{2}}{c} = \frac{4 \sqrt{29}}{29}$$$.

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

The first latus rectum is $$$y = - \sqrt{29}$$$.

The second latus rectum is $$$y = \sqrt{29}$$$.

The endpoints of the first latus rectum can be found by solving the system $$$\begin{cases} 25 x^{2} - 4 y^{2} + 100 = 0 \\ y = - \sqrt{29} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the first latus rectum are $$$\left(- \frac{4}{5}, - \sqrt{29}\right)$$$, $$$\left(\frac{4}{5}, - \sqrt{29}\right)$$$.

The endpoints of the second latus rectum can be found by solving the system $$$\begin{cases} 25 x^{2} - 4 y^{2} + 100 = 0 \\ y = \sqrt{29} \end{cases}$$$ (for steps, see system of equations calculator).

The endpoints of the second latus rectum are $$$\left(- \frac{4}{5}, \sqrt{29}\right)$$$, $$$\left(\frac{4}{5}, \sqrt{29}\right)$$$.

The length of the latera recta (focal width) is $$$\frac{2 a^{2}}{b} = \frac{8}{5}$$$.

The first directrix is $$$y = k - \frac{b^{2}}{c} = - \frac{25 \sqrt{29}}{29}$$$.

The second directrix is $$$y = k + \frac{b^{2}}{c} = \frac{25 \sqrt{29}}{29}$$$.

The first asymptote is $$$y = - \frac{b}{a} \left(x - h\right) + k = - \frac{5 x}{2}$$$.

The second asymptote is $$$y = \frac{b}{a} \left(x - h\right) + k = \frac{5 x}{2}$$$.

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

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

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

y-intercepts: $$$\left(0, -5\right)$$$, $$$\left(0, 5\right)$$$

Answer

Standard form/equation: $$$\frac{y^{2}}{5^{2}} - \frac{x^{2}}{2^{2}} = 1$$$A.

Vertex form/equation: $$$\frac{y^{2}}{25} - \frac{x^{2}}{4} = 1$$$A.

General form/equation: $$$25 x^{2} - 4 y^{2} + 100 = 0$$$A.

First focus-directrix form/equation: $$$x^{2} + \left(y + \sqrt{29}\right)^{2} = \frac{29 \left(y + \frac{25 \sqrt{29}}{29}\right)^{2}}{25}$$$A.

Second focus-directrix form/equation: $$$x^{2} + \left(y - \sqrt{29}\right)^{2} = \frac{29 \left(y - \frac{25 \sqrt{29}}{29}\right)^{2}}{25}$$$A.

Graph: see the graphing calculator.

Center: $$$\left(0, 0\right)$$$A.

First focus: $$$\left(0, - \sqrt{29}\right)\approx \left(0, -5.385164807134504\right)$$$A.

Second focus: $$$\left(0, \sqrt{29}\right)\approx \left(0, 5.385164807134504\right)$$$A.

First vertex: $$$\left(0, -5\right)$$$A.

Second vertex: $$$\left(0, 5\right)$$$A.

First co-vertex: $$$\left(-2, 0\right)$$$A.

Second co-vertex: $$$\left(2, 0\right)$$$A.

Major (transverse) axis length: $$$10$$$A.

Semi-major axis length: $$$5$$$A.

Minor (conjugate) axis length: $$$4$$$A.

Semi-minor axis length: $$$2$$$A.

First latus rectum: $$$y = - \sqrt{29}\approx -5.385164807134504$$$A.

Second latus rectum: $$$y = \sqrt{29}\approx 5.385164807134504$$$A.

Endpoints of the first latus rectum: $$$\left(- \frac{4}{5}, - \sqrt{29}\right)\approx \left(-0.8, -5.385164807134504\right)$$$, $$$\left(\frac{4}{5}, - \sqrt{29}\right)\approx \left(0.8, -5.385164807134504\right)$$$A.

Endpoints of the second latus rectum: $$$\left(- \frac{4}{5}, \sqrt{29}\right)\approx \left(-0.8, 5.385164807134504\right)$$$, $$$\left(\frac{4}{5}, \sqrt{29}\right)\approx \left(0.8, 5.385164807134504\right)$$$A.

Length of the latera recta (focal width): $$$\frac{8}{5} = 1.6$$$A.

Focal parameter: $$$\frac{4 \sqrt{29}}{29}\approx 0.742781352708207$$$A.

Eccentricity: $$$\frac{\sqrt{29}}{5}\approx 1.077032961426901$$$A.

Linear eccentricity (focal distance): $$$\sqrt{29}\approx 5.385164807134504$$$A.

First directrix: $$$y = - \frac{25 \sqrt{29}}{29}\approx -4.642383454426297$$$A.

Second directrix: $$$y = \frac{25 \sqrt{29}}{29}\approx 4.642383454426297$$$A.

First asymptote: $$$y = - \frac{5 x}{2} = - 2.5 x$$$A.

Second asymptote: $$$y = \frac{5 x}{2} = 2.5 x$$$A.

x-intercepts: no x-intercepts.

y-intercepts: $$$\left(0, -5\right)$$$, $$$\left(0, 5\right)$$$A.

Domain: $$$\left(-\infty, \infty\right)$$$A.

Range: $$$\left(-\infty, -5\right] \cup \left[5, \infty\right)$$$A.


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