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Math Help - Finding point of intersection

  1. #1
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    Finding point of intersection

    The following question i need some help on:
    Find the coordinates of the points of intersection of \frac{x^2}{4}+\frac{y^2}{9}=1 and \frac{x^2}{9}+\frac{y^2}{4}=1. Show that the points of intersection are the vertices of a square.

    P.S
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Paymemoney View Post
    The following question i need some help on:
    Find the coordinates of the points of intersection of \frac{x^2}{4}+\frac{y^2}{9}=1 and \frac{x^2}{9}+\frac{y^2}{4}=1. Show that the points of intersection are the vertices of a square.

    P.S
    start by solving for y^2 in one of the equations, and then substitute.
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  3. #3
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    As you can see \frac{x^2}{4}+\frac{y^2}{9}=1 and \frac{x^2}{9}+\frac{y^2}{4}=1 have the same variables, solve it simultaneously and take only real numbers as your solution. To prove that the points are vertices of a square,

    i) Multiply the gradient of both the diagonals and show that it is -1.
    ii)Find the distance of the diagonals, they should be equal
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  4. #4
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    i have tried to do the following:

    using simultaneous equations this is what i have done but it is incorrect:

    \frac{x^2}{4}-\frac{x^2}{9}

    \frac{x^2}{9}-\frac{x^2}{4}


    \frac{5x^2}{36}-\frac{5}{36}=0

    5x^2=\frac{5}{36}*36

    x^2=\frac{180}{36}

    x^2=\frac{5}{5}

    x=\sqrt1
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  5. #5
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    Sorry. but your solution above makes no sense. Anyway here is my solution;

    let
    \frac{x^2}{4}+\frac{y^2}{9}=1 be equation (1)
    \frac{x^2}{9}+\frac{y^2}{4}=1 be equation (2)

    multiply (1) by 9;
    \frac{9x^2}{4}+y^2=9 --------(1a)
    multiply (2) by 4;
    \frac{4x^2}{9}+y^2=4---------(2a)

    (1a)-(2a);
    \frac{9x^2}{4}-\frac{4x^2}{9}=9-4
    \frac{65x^2}{36}=5
    x^2=\frac{36}{13}
    x=\pm\sqrt\frac{36}{13}

    substitute  x=\sqrt\frac{36}{13} and  x=-\sqrt\frac{36}{13} into one of the original equations, in this case (1)

    \frac{9}{13}+\frac{y^2}{9}=1
    y=\pm\sqrt\frac{36}{13}

    therefore you have
    (\sqrt\frac{36}{13},\sqrt\frac{36}{13})
    (\sqrt\frac{36}{13},-\sqrt\frac{36}{13})
    (-\sqrt\frac{36}{13},\sqrt\frac{36}{13})
    (-\sqrt\frac{36}{13},-\sqrt\frac{36}{13})

    as the solution/coordinates of vertices

    and as you can see, they are all equidistant from the origin and that by itself shows that a quadrilateral formed by joining the points would be a square
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