Find the intersection of the two ellipses:

$\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1$

and

$\displaystyle \frac{x^2}{4} + \frac{y^2}{9} = 1$

:confused:

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- Feb 3rd 2007, 03:15 AM^_^Engineer_Adam^_^intersection of the two ellipses
Find the intersection of the two ellipses:

$\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1$

and

$\displaystyle \frac{x^2}{4} + \frac{y^2}{9} = 1$

:confused: - Feb 3rd 2007, 04:06 AMticbol
(x^2)/9 +(y^2)/4 = 1 -------------(1)

(x^2)/4 +(y^2)/9 = 1 -------------(2)

At the intersection points, the x of one ellipse is the same as the x of the other ellipse. Likewise with the y of both ellipse. So, (1) = (2) at the their intersection.

(x^2)/9 +(y^2)/4 = (x^2)/4 +(y^2)/9

(x^2)(1/9 -1/4) = (y^2)(1/9 -1/4)

x^2 = y^2

x = y -------------------------------(i)

Substitute that into, say, (1),

(x^2)/9 +(x^2)/4 = 1

(x^2)(1/9 +1/4) = 1

13x^2 = 36

x^2 = 36/13

x = +,-6/sqrt(13)

Or,

x = y = +,-(6/13)sqrt(13).

If you'd draw the two ellipses on the same xy plane, you'd see that there are 4 intersection points. Therefore, the intersection points are:

((6/13)sqrt(13),(6/13)sqrt(13)) -----1

((6/13)sqrt(13),-(6/13)sqrt(13)) ----2

(-(6/13)sqrt(13),(6/13)sqrt(13)) ----3

(-(6/13)sqrt(13),-(6/13)sqrt(13)) ---4