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A hyperbola centered at the origin has vertices at
(0,+-sqrt6) and foci at
(0,+-sqrt74).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at $(0, \pm \sqrt{6})$ and foci at $(0, \pm \sqrt{74})$ . $\newline$ Write the equation of this hyperbola.

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Write equations of hyperbolas in standard form using properties

Full solution

Q. A hyperbola centered at the origin has vertices at

(0, \pm \sqrt{6})

and foci at

(0, \pm \sqrt{74})

.

\newline

Write the equation of this hyperbola.

Standard Form of Hyperbola Equation: Identify the standard form of the equation for a hyperbola with a vertical transverse axis. $\newline$ Standard form of equation for a hyperbola with a vertical transverse axis: $\newline$ $(y-k)^2/a^2 - (x-h)^2/b^2 = 1$
Determining the Center: Determine the center $(h, k)$ of the hyperbola. Since the hyperbola is centered at the origin, $h = 0$ and $k = 0$ .
Finding the Semi-Major Axis: Find the value of the semi-major axis $a$ . The vertices are at $(0, \pm\sqrt{6})$ , so $a = \sqrt{6}$ .
Calculating the Distance to the Foci: Calculate the distance $c$ between the center and the foci. The foci are at $(0, \pm\sqrt{74})$ , so $c = \sqrt{74}$ .
Using the Relationship between a, b, and c: Use the relationship $c^2 = a^2 + b^2$ to find the value of $b$ . Substitute the known values of $a$ and $c$ into the equation. $c^2 = a^2 + b^2$ $\sqrt{74}^2 = \sqrt{6}^2 + b^2$ $74 = 6 + b^2$ $b^2 = 74 - 6$ $b^2 = 68$ $b = \sqrt{68}$
Writing the Equation in Standard Form: Write the equation of the hyperbola in standard form after substituting the values of $h$ , $k$ , $a$ , and $b$ . Substitute values of $h$ , $k$ , $a$ , and $b$ into $(y-k)^2/a^2 - (x-h)^2/b^2 = 1$ . $(y - 0)^2/(\sqrt{6})^2 - (x - 0)^2/(\sqrt{68})^2 = 1$ $k$ $0$

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