# Thread: Finding point of intersection

1. ## 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

2. Originally Posted by Paymemoney
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.

3. 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

4. i have tried to do the following:

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

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

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

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

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

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

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

$\displaystyle x=\sqrt1$

5. 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