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If a straight line passing through the center of the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1 intersects the ellipse at two points a and B, and F is a focal point of the ellipse, what is the maximum area of △ ABF Such as the title Why are a and B symmetrical about o. Not quite understand
You are Fu Ren's
Let the ordinates of a and B be y a, y B, a and B symmetric with respect to point O, so | y a | = | y B |,
We know that a = 5, B = 4, C = 3, so | ya | = | Yb | ≤ 4,
So s △ ABF = 1 / 2 * | of | * | - ya | + 1 / 2 * | - of | * | - Yb | = C | - ya | ≤ 3 * 4 = 12
The intersection of line L: X-Y = 0 and ellipse xsquare / 2 + ysquare = 1 and AB two points c are the moving points on the ellipse to find the maximum area of triangle ABC
Can this problem be solved without parametric equation? Can it be solved according to the formula for the distance between a point and a straight line? I can't work it out for a long time = = I don't use parametric equation very often
Simultaneous line L: X-Y = 0 and ellipse x squared / 2 + y squared = 1 get a (√ 6 / 3, √ 6 / 3) B (- √ 6 / 3, - √ 6 / 3) | ab | = 4 √ 3 / 3 with line AB as the bottom and point C as the vertex. The maximum area of triangle ABC is the maximum distance from point C to ab. if no parametric equation is used, a straight line is required to be level with line L
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