# intersection of the two ellipses

• Feb 3rd 2007, 03:15 AM
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 AM
ticbol
Quote:

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:

(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