anonymous
  • anonymous
what is the standard form of the hyperbola with foci at (0,+-5) and vertices at (0,+-2) ?
Mathematics
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I don't remember the equations for hyperbolas foci and vertices. Only that \(x^2-y^2=r^2\) is a hyperbola.
anonymous
  • anonymous
Hey Jhanny you are awake!
Jhannybean
  • Jhannybean
standard form of hyperbola left and right is \[\large \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] First we need to find out our center using our foci and our vertices. graphing our foci and our vertices, we can count the distance between them to find the center for each. foci: \(\color{blue} {\large 2c=10 }\implies \large\color{blue}{c=5}\) from (0,5) to (0,-5) the center of our foci is 5 units, meaning the center of our foci is located at (0,0). verticles \(\color{red}{\large 2a=4 \implies a=2}\)from (0,2) to (0,-2) our distance is 2, making the center of our vertices (0,0) So the center of our hyperbola is :\(\color{purple}{\large (h,k) \implies (0,0)}\) in order to get our "b" we can use the pythagorean theroem, \[\large c^2=a^2+b^2\]\[\large 25=4+b^2\]\[\color{green}{\large b^2=21}\] Now we can write the equation of our hyperbola in standard form :) \[\large \frac{(x-\color{purple}0)^2}{\color{red}{4}}-\frac{(y-\color{purple}0)^2}{\color{green}{21}}=1\]\[\large \frac{x^2}{4}-\frac{y^2}{21}=1\]

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sirm3d
  • sirm3d
the foci lie on the y-axis, so the correct equation of the hyperbola should be \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\]
Jhannybean
  • Jhannybean
Ah, then...\[\large \frac{y^2}{21}-\frac{x^2}{4}=1\]***** I'm still learning xD. Thank you @sirm3d

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