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[1/1 Points]
DETAILS
MY NOTES
Find the standard form of the equation of the hyperbola with the given characteristics.
Vertices:
(+-6,0); foci:
(+-9,0)

$2$ . [ $1$ / $1$ Points] $\newline$ DETAILS $\newline$ MY NOTES $\newline$ Find the standard form of the equation of the hyperbola with the given characteristics. $\newline$ Vertices: $( \pm 6,0)$ ; foci: $( \pm 9,0)$

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Convert complex numbers between rectangular and polar form

Full solution

Q.

2

. [

1

/

1

Points]

\newline

DETAILS

\newline

MY NOTES

\newline

Find the standard form of the equation of the hyperbola with the given characteristics.

\newline

Vertices:

( \pm 6,0)

; foci:

( \pm 9,0)

Calculate a^ $2$ : The standard form of the equation of a hyperbola with a horizontal transverse axis centered at the origin is given by: $\newline$ $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $\newline$ where $2a$ is the distance between the vertices, and $2c$ is the distance between the foci. We are given the vertices at (± $6$ , $0$ ), which means $a = 6$ . We can calculate $a^2$ as follows: $\newline$ $a^2 = 6^2 = 36$
Calculate c^ $2$ : Next, we are given the foci at (± $9$ , $0$ ), which means $c = 9$ . We can calculate $c^2$ as follows: $\newline$ $c^2 = 9^2 = 81$
Find b^ $2$ : We know that for a hyperbola, the relationship between $a$ , $b$ , and $c$ is given by $c^2 = a^2 + b^2$ . We can use this to find $b^2$ : $\newline$ $b^2 = c^2 - a^2 = 81 - 36 = 45$
Write standard form: Now that we have $a^2$ and $b^2$ , we can write the standard form of the equation of the hyperbola: $\newline$ $\frac{x^2}{36} - \frac{y^2}{45} = 1$

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