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K.Binks

  • one year ago

Write the equation of a hyperbola with vertices (-5, 0) and (5, 0) and co-vertices at (0, -6) and (0, 6)?

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  1. jdoe0001
    • one year ago
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    |dw:1438211364085:dw| get the "a" and "b" components firstly then plug them in keep in mind that, the vertices are at 5,0 and -5,0, so it opens horizontally, meaning the fraction with the "x" variable has the "a" component, that is \(\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ a}}^2}-\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ b}}^2}=1 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad vertices\ ({\color{brown}{ h}}\pm a, {\color{blue}{ k}})\)

  2. K.Binks
    • one year ago
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    So... x^2/25 - y^2/36 =1? My biggest issue doing these problems is identifying the A and B, H and K. once I know those i can plug them in. I think I got this one, but mostly cause I narrowed down the answer choices.. how do you tell which part is which?

  3. jdoe0001
    • one year ago
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    hmmm which part is "a" or "b"? in the case of ellipses, the "bigger" denominator is "a" all the time in a hyperbola, "a" can be smaller than "b", BUT "a" is always under the variable whose axis the hyperbola opens up so if the hyperbola opens up horizontally, it opens in the direction of the x-axis that means, that the fraction with the "x" is positive, and "a" is under it if the hyperbola opens up vertically, means it opens in the direction of the y-axis that means the fraction with the "y" is the positive one, and the "a" is under it

  4. jdoe0001
    • one year ago
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    x^2/25 - y^2/36 =1 \(\Large \checkmark\)

  5. K.Binks
    • one year ago
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    THANK YOU! That cleared up so much for me, thanks!!

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