You are not logged in.
Pages: 1
Hello.
Part of a question I was doing required me to find the points of intersection of two ellipses. I'm a bit confused about one of the answers.
The two ellipses are: (x+2)² +2y² =18 and 9(x-1)² +16y² =25
I multiplied the first by 8, eliminated y and solved for x. I end up with a quadratic x² - 50x + 96 = 0, which I factorise to give (x-2)(x-48). But substituting x=48 into either original equation will give a y value that is the square root of a negative number. I assume that this solution isn't valid, and that the two points of intersection of the ellipses both have an x-coordinate of 2. However, I was wondering where the 48 came from? All I've done is multiply by 8, which shouldn't create an extra solution?? Can anyone explain?
Thanks.
Offline
Thats because x = 48 IS a solution to the two equations, treated as joint-quadratic equations in x and y rather than as equations of ellipses.
When x = 48, y = ±1241i.
Of course, if you want only solutions that give the intersections of the ellipses, then you have to ignore complex values; hence x = 48 would have to be rejected.
Offline
Ahh that makes sense, thank you Jane.
Offline
Pages: 1