Let AB be the chord of the center of the ellipse x ^ 2 / 9 + y ^ 2 / 25 = 1, and f be a focal point of the ellipse, then the maximum area of the triangle ABF is

Let AB be the chord of the center of the ellipse x ^ 2 / 9 + y ^ 2 / 25 = 1, and f be a focal point of the ellipse, then the maximum area of the triangle ABF is

Let the center of the ellipse be the coordinate origin o (0,0). From the symmetry of the ellipse, we can see that the distance from the chord AB passing through point O to the upper and lower focus is equal. Therefore, Let f be the upper focus, that is, f (0,4)
The ellipse is symmetric about y axis, so the distance from point a to y axis = the distance from point B to y axis
so
S△ABF=S△AOF+S△BOF
=Distance from 1 / 2 * of * point a to y axis + distance from 1 / 2 * of * point B to y axis
=Of * distance from a to y axis
≤4×3=12
The maximum area of the triangle ABF is 12
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