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Find an equation in standard form for the hyperbola with vertices at (0, ±6) and asymptotes at y=+-3/4x?
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The vertices are on the y axis so the equation has this form: \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]
a=6
The asymptotes are: \[y=\frac{\pm a}{b}x\]
@Mertsj thank you!
yw
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So \[\frac{6}{b}=\frac{3}{4}\]
@Mertsj okay what is the next step?
What is b?
b=8
What is a^2?
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a^2=36
What is b^2?
b^2=64
Write the equation.
so its y^2/36-x^2/64=1?
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yes
@Mertsj thanks again!
yw
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