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Math Help - Equation Questions

  1. #1
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    Equation Questions

    1.If I have the equation (x+4)squared + (y+4)squared = 9

    The center of the circle would be (-4,-4) correct?

    2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)

    The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?

    Is there an easier way to write this equation?

    Thanks.
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  2. #2
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    Okay, I figured out the first part. Do any of you know number two.
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  3. #3
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    Quote Originally Posted by milosh View Post
    1.If I have the equation (x+4)squared + (y+4)squared = 9

    The center of the circle would be (-4,-4) correct?

    correct

    2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)

    The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?

    Is there an easier way to write this equation?

    yes, there is ...

    \textcolor{red}{4(x-20)^2 + 16(y-10)^2 = 1}
    ...
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  4. #4
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    Quote Originally Posted by milosh View Post
    1.If I have the equation (x+4)squared + (y+4)squared = 9

    The center of the circle would be (-4,-4) correct?

    2. If I have an ellipse with the equation (x-20)squared over 1/4 + (y-10)squared over 1/16 (over is the division sign)
    Do you mean "over 1/4" or "over 4"? If you actually mean "over 1/4" and "over 1/16", the equation is
    \frac{(x-20)^2}{1/4}+ \frac{(y-10)^2}{1/16}= 4(x-20)^2+ 16(y-10)^2= 1.

    If you meant "over 4" and "over 16", the equation is
    \frac{(x-20)^2}{4}+ \frac{(y-10)^2}{16}= 1
    which can also be written as
    16(x- 20)^2+ 4(y-10)^2= 64
    after multiplying both sides by 4(16)= 64.

    The 1/4 is truly the distance of 1/2 and the 1/16 is truly the distance of 1/4 correct?
    That sentence doesn't make sense. If you mean "1/4 is the length of the semi-axis parallel to the x-axis and 1/16 is the length of the semi-axis parallel to the y-axis" and you really are talking about
    \frac{(x-20)^2}{1/4}+ \frac{(y-10)^2}{1/16}= 4(x-20)^2+ 16(y-10)^2= 1
    then the answer is "No". The lengths of the x and y semi-axes are 1/\sqrt{4}= 1/2 and 1/\sqrt{16}= 1/4 respectively. If you actually meant
    16(x- 20)^2+ 4(y-10)^2= 64
    then the answer is stilll "No"- the lengths of the x and y semi axes are \sqrt{4}= 2 and \sqrt{16}= 4, respectively,

    Is there an easier way to write this equation?

    Thanks.
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