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A hyperbola centered at the origin has vertices at
(+-sqrt33,0) and foci at
(+-sqrt59,0).
Write the equation of this hyperbola.

A hyperbola centered at the origin has vertices at $( \pm \sqrt{33}, 0)$ and foci at $( \pm \sqrt{59}, 0)$ . $\newline$ Write the equation of this hyperbola.

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Convert equations of conic sections from general to standard form

Full solution

Q. A hyperbola centered at the origin has vertices at

( \pm \sqrt{33}, 0)

and foci at

( \pm \sqrt{59}, 0)

.

\newline

Write the equation of this hyperbola.

Identify Hyperbola Equation: The equation of a hyperbola centered at the origin with horizontal transverse axis is given by $(\frac{x^2}{a^2}) - (\frac{y^2}{b^2}) = 1$ , where $2a$ is the distance between the vertices and $2c$ is the distance between the foci. We are given the vertices at $(\pm\sqrt{33}, 0)$ , so $a = \sqrt{33}$ .
Find Distance Between Vertices and Foci: Next, we are given the foci at $(\pm\sqrt{59}, 0)$ , so $c = \sqrt{59}$ . The relationship between $a$ , $b$ , and $c$ in a hyperbola is $c^2 = a^2 + b^2$ . We can use this to solve for $b^2$ .
Calculate $b^2$ : Substitute the known values of $a$ and $c$ into the relationship $c^2 = a^2 + b^2$ to find $b^2$ . $\newline$ $(\sqrt{59})^2 = (\sqrt{33})^2 + b^2$ $\newline$ $59 = 33 + b^2$ $\newline$ $b^2 = 59 - 33$ $\newline$ $b^2 = 26$
Write Hyperbola Equation: Now that we have $a^2$ and $b^2$ , we can write the equation of the hyperbola. Substitute $a^2 = 33$ and $b^2 = 26$ into the standard equation of the hyperbola. $(\frac{x^2}{33}) - (\frac{y^2}{26}) = 1$

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