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A hyperbola centered at the origin has vertices at
(0,+-sqrt21) and foci at
(0,+-sqrt34).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at $(0, \pm \sqrt{21})$ and foci at $(0, \pm \sqrt{34})$ . $\newline$ 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{21})

and foci at

(0, \pm \sqrt{34})

.

\newline

Write the equation of this hyperbola.

Identify standard form: Identify the standard form of the equation for a hyperbola centered at the origin with a vertical transverse axis. $\newline$ Standard form of equation for a hyperbola with vertical transverse axis: $\newline$ $(y-k)^2/a^2 - (x-h)^2/b^2 = 1$ where $(h, k)$ is the center of the hyperbola.
Determine center coordinates: Determine the values of $h$ and $k$ for the center of the hyperbola. Since the hyperbola is centered at the origin, $h = 0$ and $k = 0$ .
Find semi-major axis: Find the value of the semi-major axis $a$ . The vertices are at $(0, \pm\sqrt{21})$ , so the distance from the center to a vertex is $a = \sqrt{21}$ .
Find semi-minor axis: Find the value of the semi-minor axis $b$ . The relationship between the semi-major axis $a$ , the semi-minor axis $b$ , and the distance to the foci $c$ for a hyperbola is $c^2 = a^2 + b^2$ . The foci are at $(0, \pm\sqrt{34})$ , so $c = \sqrt{34}$ .
Calculate $b$ using $c^2 = a^2 + b^2$ : Calculate the value of $b$ using the relationship $c^2 = a^2 + b^2$ .
$c^2 = (\sqrt{34})^2$
$c^2 = 34$
$a^2 = (\sqrt{21})^2$
$a^2 = 21$
Substitute $a^2$ and $c^2$ into the relationship to find $c^2 = a^2 + b^2$ $0$ .
$c^2 = a^2 + b^2$ $1$
$c^2 = a^2 + b^2$ $2$
$c^2 = a^2 + b^2$ $3$
Write equation in standard form: Write the equation of the hyperbola in standard form after substituting the values of $h$ , $k$ , $a$ , and $b$ . Substitute $h = 0$ , $k = 0$ , $a^2 = 21$ , and $b^2 = 13$ into $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ . $\frac{(y - 0)^2}{21} - \frac{(x - 0)^2}{13} = 1$ $k$ $0$

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