Finding point of intersection

**Paymemoney** · January 16th 2010, 09:13 PM

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

**VonNemo19** · January 16th 2010, 10:34 PM

Originally Posted by Paymemoney

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.

**bryankek** · January 16th 2010, 10:42 PM

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

**Paymemoney** · January 17th 2010, 01:12 AM

i have tried to do the following:

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

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

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

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

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

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

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

$x=\sqrt1$

**bryankek** · January 17th 2010, 03:31 AM

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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