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Thread: circle belonging to a ellipse

  1. #1
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    circle belonging to a ellipse

    See attachment, thanks for looking
    Attached Thumbnails Attached Thumbnails circle belonging to a ellipse-untitled.jpg  
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  2. #2
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    The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $\displaystyle F_1P + F_2P = c$. Then the equation of the ellipse is $\displaystyle \sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$. I simplified this equation to $\displaystyle \frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$. Now can you show that the circle $\displaystyle x^2 + y^2 = \frac{c^2}{4}$ has the desired property?
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $\displaystyle F_1P + F_2P = c$. Then the equation of the ellipse is $\displaystyle \sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$. I simplified this equation to $\displaystyle \frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$. Now can you show that the circle $\displaystyle x^2 + y^2 = \frac{c^2}{4}$ has the desired property?
    How did u simplify your the equation to the following $\displaystyle \frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$.??? I tired to square both sides but it just leaves one big mess and i can never get down to the same equation you got.
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  4. #4
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    Ok, here goes:

    $\displaystyle \sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$

    $\displaystyle \sqrt{y^2 + (x+d)^2} = c - \sqrt{y^2 + (x-d)^2}$

    $\displaystyle y^2 + (x+d)^2 = c^2 - 2c\sqrt{y^2 + (x-d)^2} + y^2 + (x-d)^2$

    $\displaystyle (x+d)^2 - (x-d)^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

    $\displaystyle x^2 + 2xd + d^2 - (x^2 - 2xd + d^2) - c^2 = -2c\sqrt
    {y^2 + (x-d)^2}$

    $\displaystyle x^2 + 2xd + d^2 - x^2 + 2xd - d^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

    $\displaystyle 4xd - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

    $\displaystyle (4xd - c^2)^2 = 4c^2(y^2 + (x-d)^2)$

    $\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x-d)^2$

    $\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x^2 - 2xd + d^2)$

    $\displaystyle 16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2x^2 - 8xdc^2 + 4c^2d^2$

    $\displaystyle 16x^2d^2 + c^4 = 4c^2y^2 + 4c^2x^2 + 4c^2d^2$

    $\displaystyle (16d^2 - 4c^2)x^2 - 4c^2y^2 = 4c^2d^2 - c^4$

    $\displaystyle 4(4d^2 - c^2)x^2 - 4c^2y^2 = c^2(4d^2 - c^2)$

    $\displaystyle \frac{4x^2}{c^2} - \frac{4y^2}{4d^2 - c^2} = 1$

    $\displaystyle \frac{4x^2}{c^2} + \frac{4y^2}{c^2 - 4d^2} = 1$
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  5. #5
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    i know i need to confirm that substituting into the ellipse and into the circle will result in the same values of x

    this k is equal to . How does one know that k is equal to this??

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  6. #6
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    Also, have another question. How does one know that the center of the circle in this problem is at c^2/4??
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