Math / Geometry

QuestionQuestion 5, 8.2.71
Write an equation for the hyperbola that has foci at $(0, \sqrt{5})$ and $(0,-\sqrt{5})$ , and asymptotes $y= \pm 3 x$ .
An equation for the hyperbola is $\square$ (Simplify your answer. Type your answer in standard form. Use integers or fractions for any numbers in the equation.)

Studdy Solution

STEP 1

1. The foci of the hyperbola are located at $(0, \sqrt{5})$ and $(0, -\sqrt{5})$ .
2. The asymptotes of the hyperbola are given by the equations $y = \pm 3x$ .
3. The hyperbola is centered at the origin $(0, 0)$ .

STEP 2

1. Determine the orientation of the hyperbola.
2. Identify the values of $a$ , $b$ , and $c$ using the given information.
3. Write the equation of the hyperbola in standard form.

STEP 3

Determine the orientation of the hyperbola:
Since the foci are on the y-axis, the hyperbola is vertical. The standard form of a vertical hyperbola is:
$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

STEP 4

Identify the values of $a$ , $b$ , and $c$ :
- The distance from the center to each focus is $c = \sqrt{5}$ . - The slopes of the asymptotes are $\pm \frac{a}{b} = \pm 3$ .
Using the relationship $c^2 = a^2 + b^2$ , we have:
$c = \sqrt{5} \Rightarrow c^2 = 5$
From the asymptotes, $\frac{a}{b} = 3$ , we can express $a$ in terms of $b$ :
$a = 3b$
Substitute $a = 3b$ into the equation for $c^2$ :
$5 = (3b)^2 + b^2$ $5 = 9b^2 + b^2$ $5 = 10b^2$
Solve for $b^2$ :
$b^2 = \frac{5}{10} = \frac{1}{2}$
Now, find $a^2$ using $a = 3b$ :
$a^2 = (3b)^2 = 9b^2 = 9 \times \frac{1}{2} = \frac{9}{2}$

STEP 5

Write the equation of the hyperbola in standard form:
Substitute $a^2$ and $b^2$ into the standard form of the vertical hyperbola:
$\frac{y^2}{\frac{9}{2}} - \frac{x^2}{\frac{1}{2}} = 1$
Multiply through by 2 to simplify:
$\frac{2y^2}{9} - 2x^2 = 1$
Multiply through by 9 to eliminate fractions:
$2y^2 - 9x^2 = 9$
The equation of the hyperbola is:
$\boxed{2y^2 - 9x^2 = 9}$

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