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Thread: 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 $\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$ and $\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$. 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 $\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$ and $\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$. Show that the points of intersection are the vertices of a square.

    P.S
    start by solving for $\displaystyle y^2$ in one of the equations, and then substitute.
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  3. #3
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    As you can see $\displaystyle \frac{x^2}{4}+\frac{y^2}{9}=1$ and $\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$ 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:

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

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


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

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

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

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

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

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

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

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

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

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

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

    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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