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Math Help - One yes or no and one ellipse equation

  1. #1
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    One yes or no and one ellipse equation

    For ((X^2)/ 9) + ((Y^2)/16) = 1

    the general form after 4 units right and 2 units down would be:
    16x^2 + 9y^2 - 128x +36y + 148 = 0 Yes/No

    and for the equation (((x-2)^2)\4) + (((y-3)^2)/16) = 1

    the standard equation would be?
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  2. #2
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    Quote Originally Posted by Stuart
    For ((X^2)/ 9) + ((Y^2)/16) = 1

    the general form after 4 units right and 2 units down would be:
    16x^2 + 9y^2 - 128x +36y + 148 = 0 Yes/No

    and for the equation (((x-2)^2)\4) + (((y-3)^2)/16) = 1

    the standard equation would be?
    The isometry here is a translantion,
    \tau: (x,y)\to (x-4,y-2)
    Thus,
    \frac{x^2}{9}+\frac{y^2}{16}=1
    Becomes,
    \frac{(x-4)^2}{9}+\frac{(y-2)^2}{16}=1
    Thus,
    16(x-4)^2+9(y-2)^2=144
    Thus,
    16x^2-128x+256+9y^2-36y+36=144
    Thus,
    16x^2-128x+9y^2-36y+148=0
    Your mistake it that you used the translation,
    \tau: (x,y)\to (x-4,y+2)
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  3. #3
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    ok so how would I start the second equation listed?
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  4. #4
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    Quote Originally Posted by Stuart
    ok so how would I start the second equation listed?
    By the previous post you have,
    \frac{(x-2-4)^2}{9}+\frac{(y-3-2)^2}{16}=1
    Thus,
    \frac{(x-6)^2}{9}+\frac{(y-5)^2}{16}=1
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