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assignment 7.5: hyperbolas
score: 7/15 answered: 6/15

question 7
find the standard form equation for a hyperbola with vertices at $(9, 0)$ and $(-9, 0)$ and asymptote $y = \frac{1}{3}x$

Explanation:

Step1: Identify hyperbola orientation

Vertices $(\pm9,0)$ lie on x-axis, so horizontal hyperbola. Standard form: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Step2: Find value of $a$

Vertex distance from origin: $a = 9$, so $a^2 = 81$

Step3: Use asymptote to find $b$

Asymptote for horizontal hyperbola: $y = \frac{b}{a}x$. Given $y = \frac{1}{3}x$, so $\frac{b}{a} = \frac{1}{3}$. Substitute $a=9$:
$\frac{b}{9} = \frac{1}{3} \implies b = 3$, so $b^2 = 9$

Step4: Substitute into standard form

Plug $a^2=81$ and $b^2=9$ into the equation.

Answer:

$\frac{x^2}{81} - \frac{y^2}{9} = 1$