SOLUTION: The equation of an ellipse with center (5,3) that passes through the points (9,3) and (5,5) has the form f(x,y)=1. Find f(x,0).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The equation of an ellipse with center (5,3) that passes through the points (9,3) and (5,5) has the form f(x,y)=1. Find f(x,0).       Log On


   



Question 822959: The equation of an ellipse with center (5,3) that passes through the points (9,3) and (5,5) has the form f(x,y)=1. Find f(x,0).

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This problem is not very difficult once you recognize that the two points, (9, 3) and (5, 5), are not just any two points on the ellipse. Since (9, 3) has the same y-coordinate as the center it is a vertex on the horizontal axis. And since (5, 5) has the same x-coordinate as the center, it is a vertex on the vertical axis.

Since the distance from (9, 3) to the center is 6 and the distance from (5, 5) to the center is 2: a = 6, b = 2 and the ellipse is horizontal. With a center of (5, 3), the equation for the ellipse is:
%28x-5%29%5E2%2F6%5E2+%2B+%28y-3%29%5E2%2F2%5E2+=+1
The left side the the f(x, y) mentioned in the problem. To find f(x, 0) we just replace the "y" with zero in f(x, y):
%28x-5%29%5E2%2F6%5E2+%2B+%280-3%29%5E2%2F2%5E2
Now we simplify:
%28x-5%29%5E2%2F6%5E2+%2B+%28-3%29%5E2%2F2%5E2
%28x%5E2-10x%2B25%29%2F36+%2B+9%2F4
%28x%5E2-10x%2B25%29%2F36+%2B+%289%2F4%29%289%2F9%29
%28x%5E2-10x%2B25%29%2F36+%2B+81%2F36
%28x%5E2-10x%2B106%29%2F36
This is f(x, 0).