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Thread: hyperbola

  1. #1
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    hyperbola

    whats the standard form of this hyperbola 4x^2 - 25y^2 - 50y - 125= 0
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    Junior Member ursa's Avatar
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    whats the standard form of this hyperbola 4x^2 - 25y^2 - 50y - 125= 0
    (x^2/25)-((y+1)^2/4)=1
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    Quote Originally Posted by 14041471 View Post
    whats the standard form of this hyperbola 4x^2 - 25y^2 - 50y - 125= 0
    The standard form should have been mentioned in your book and / or class lecture!

    To arrange the given hyperbola in the form:

    . . . . .$\displaystyle \frac{(x\, -\, h)^2}{a^2}\, -\, \frac{(y\, -\, k)^2}{b^2}\, =\, 1$

    ...you will need to complete the one square (since $\displaystyle 4x^2$ is obviously just $\displaystyle 4(x\, -\, 0)^2$), and then divide through by whatever number you end up with on the right-hand (non-variable) side of the equation.

    If you're unclear on the steps in this process, first study two or more lessons online to review. Then attempt this exercise. If you get stuck, you will then be able to reply with a clear listing of your steps and reasoning so far, which will enable helpers to provide you with this most-effective assistance.

    Thank you!
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    Quote Originally Posted by ursa View Post
    (x^2/25)-((y+1)^2/4)=1
    Instead of just stating an answer, It's generally a good idea to give some idea of where the answer comes. eg. a suggestion to complete the square and perhaps the first line of the calculation .... See the previous post on one way of doing this.
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