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Math Help - Intersection of Circle and Parabola

  1. #1
    Senior Member polymerase's Avatar
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    Intersection of Circle and Parabola

    A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola y=2x^2 tangentially at exactly 2 points. Find these 2 points.

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola y=2x^2 tangentially at exactly 2 points. Find these 2 points.

    Thanks!
    Let the center of the circle be (0,a) for a>0

    thus our two curves are:

    x^2 + (y - a)^2 = 1 and y = 2x^2

    and we want these two curves to have only two points in common. (it is okay to take this interpretation since the circle will be "inside" the parabola)

    plug in y = 2x^2 into the equation of the circle, we get:

    x^2 + \left( 2x^2 - a \right)^2 = 1

    \Rightarrow 4x^4 + (1 - 4a)x^2 + \left( a^2 - 1 \right) = 0

    the above is quadratic in x^2

    thus, x^2 = \frac {4a - 1 \pm \sqrt{(1 - 4a)^2 - 16 \left( a^2 - 1 \right) }}8 ..........................(1)

    we want this to have only two solutions, therefore, set the discriminant to zero.

    thus, it remains for us to solve:

    (1 - 4a)^2 - 16 \left( a^2 - 1 \right) = 0 for a.

    knowing the value for a, we plug it into equation (1) to find the x-coordinate for the point of intersection. then use that to get the y-coordinate. (note, one of your solutions for y will be erroneous, be sure to make sure your solutions work in BOTH of the original equations). i leave the rest to you, it's pretty much following your nose from here
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by polymerase View Post
    A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola y=2x^2 tangentially at exactly 2 points. Find these 2 points.

    Thanks!
    i have solved this for my tutor before but i can't find my solutions..
    but i think, this can be solved in two ways, one is using calculus while the other is using some college algebra..

    let (0,k), k>0 be the center of the circle: x^2 + (y-k)^2 = 1
    y = 2x^2

    let us use the fact that both are symmetric wrt to y axis.. hence, let (a,b) and (-a,b) be the points of intersection..

    so, a^2 + (b-k)^2 = 1 and b = 2a^2 and so if we combine both, we have
    a^2 +(2a^2 - k)^2 = 4a^4 - a^2(4k - 1) + k^2 = 1

    or 4a^4 - a^2(4k - 1) + (k^2 - 1) = 0

    using quadratic formula, we have:

    a^2 = \frac{4k-1 \pm \sqrt{(4k - 1)^2 - 16(k^2-1)}}{8} = \frac{4k-1 \pm \sqrt{- 8k + 17}}{8}

    for that to exists, we need - 8k + 17 \geq 0 or k \leq \frac{17}{8}

    in fact, what we need is that k=\frac{17}{8} so that we have a unique solution for a^2 and that a has 2 solutions.. Ü so if k=\frac{17}{8}, we have a^2 = \frac{60}{64} or a= \pm \frac{\sqrt{60}}{8}. and b=\frac{60}{32}

    hence, the points are \left( {\frac{\sqrt{60}}{8}, \frac{60}{32}} \right) and \left( {\frac{-\sqrt{60}}{8}, \frac{60}{32}} \right)

    EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kalagota View Post
    EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe
    hehe, sorry about that. you can do the calculus way if you want. the poster may want to see that since he/she is in calculus
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Jhevon View Post
    Let the center of the circle be (0,a) for a>0

    thus our two curves are:

    x^2 + (y - a)^2 = 1 and y = 2x^2

    and we want these two curves to have only two points in common. (it is okay to take this interpretation since the circle will be "inside" the parabola)

    plug in y = 2x^2 into the equation of the circle, we get:

    x^2 + \left( 2x^2 - a \right)^2 = 1

    \Rightarrow 4x^4 + (1 - 4a)x^2 + \left( a^2 - 1 \right) = 0

    the above is quadratic in x^2

    thus, x^2 = \frac {4a - 1 \pm \sqrt{(1 - 4a)^2 - 16 \left( a^2 - 1 \right) }}8 ..........................(1)

    we want this to have only two solutions, therefore, set the discriminant to zero.

    thus, it remains for us to solve:

    (1 - 4a)^2 - 16 \left( a^2 - 1 \right) = 0 for a.

    knowing the value for a, we plug it into equation (1) to find the x-coordinate for the point of intersection. then use that to get the y-coordinate. (note, one of your solutions for y will be erroneous, be sure to make sure your solutions work in BOTH of the original equations). i leave the rest to you, it's pretty much following your nose from here
    Thanks for the relay!
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  6. #6
    Senior Member polymerase's Avatar
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    Quote Originally Posted by kalagota View Post
    i have solved this for my tutor before but i can't find my solutions..
    but i think, this can be solved in two ways, one is using calculus while the other is using some college algebra..

    let (0,k), k>0 be the center of the circle: x^2 + (y-k)^2 = 1
    y = 2x^2

    let us use the fact that both are symmetric wrt to y axis.. hence, let (a,b) and (-a,b) be the points of intersection..

    so, a^2 + (b-k)^2 = 1 and b = 2a^2 and so if we combine both, we have
    a^2 +(2a^2 - k)^2 = 4a^4 - a^2(4k - 1) + k^2 = 1

    or 4a^4 - a^2(4k - 1) + (k^2 - 1) = 0

    using quadratic formula, we have:

    a^2 = \frac{4k-1 \pm \sqrt{(4k - 1)^2 - 16(k^2-1)}}{8} = \frac{4k-1 \pm \sqrt{- 8k + 17}}{8}

    for that to exists, we need - 8k + 17 \geq 0 or k \leq \frac{17}{8}

    in fact, what we need is that k=\frac{17}{8} so that we have a unique solution for a^2 and that a has 2 solutions.. Ü so if k=\frac{17}{8}, we have a^2 = \frac{60}{64} or a= \pm \frac{\sqrt{60}}{8}. and b=\frac{60}{32}

    hence, the points are \left( {\frac{\sqrt{60}}{8}, \frac{60}{32}} \right) and \left( {\frac{-\sqrt{60}}{8}, \frac{60}{32}} \right)

    EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe
    Thanks!
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