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dricboylan

  • 2 years ago

What are the vertices of the hyperbola given by the equation (y-4)^2/121 - (x+9)^2/141 = 1?

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  1. satellite73
    • 2 years ago
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    first you need the center so you know that one ?

  2. dricboylan
    • 2 years ago
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    No, I really don't understand this section which is why I was asking for help.

  3. satellite73
    • 2 years ago
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    center of \[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\]is \((h,k)\)

  4. satellite73
    • 2 years ago
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    in your case it is \((-9,4)\)

  5. satellite73
    • 2 years ago
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    because the \(y\) part comes first (before the minus sign) this hyperbola looks like this (roughly) |dw:1353293873616:dw|

  6. satellite73
    • 2 years ago
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    to find the vertices, go \(a\) units up up and down from the center. that is why you needed the center first in your case \(a^2=121\) and so \(a=11\)

  7. satellite73
    • 2 years ago
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    since the center is \((-9,4)\) if you go up 11 units you are at \((-9,15)\) and if you go down 11 units you are at \((-9,-7)\)

  8. satellite73
    • 2 years ago
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    hope that is clear, and also that you see it is not that hard once you know what you are doing

  9. dricboylan
    • 2 years ago
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    I'm still extremely lost, as you can tell math isn't my strong point. I appreciate the help however.

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