A vertex on the major axis of the ellipse x2 + 4y2 = 4 is a. take a as the right vertex to make an isosceles right triangle inscribed on the ellipse. The area of the triangle is______ ．

A vertex on the major axis of the ellipse x2 + 4y2 = 4 is a. take a as the right vertex to make an isosceles right triangle inscribed on the ellipse. The area of the triangle is______ ．

Let a be (- 2, 0) so that one is y = x + 2 and substitute it into the ellipse 5x2 + 16x + 12 = 0 (5x + 6) (x + 2) = 0x = - 65, x = - 2 (excluding) x = - 65, y = x + 2 = 45, so the intersection of a and ellipse is C (- 65, 45), then ac2 = (- 2 + 65) 2 + (0-45) 2 = 3225, so the area product =

A vertex on the major axis of the ellipse x ^ 2 + 4Y ^ 2 = 4 is a. take a as the right vertex, make an isosceles right triangle inscribed with the ellipse, and calculate the area?

X ^ 2 / 4 + y ^ 2 = 1 let a (2,0) isosceles right triangle be symmetric about X axis, so the angle between waist and X axis is 45, so a waist is y = tan45 (X-2) = X-2, and substituting 5x ^ 2-16x + 12 = 0 (X-2) (5x-6) = 0x = 2 is a, so x = 6 / 5, y = X-2 = - 4 / 5, so another vertex is B (6 / 5,4 / 5), then the right edge AB ^ 2 =

Given the elliptic equation x ^ 2 / A ^ 2 + y ^ 2 / b ^ = 1 (a > b > 0), what is the maximum area of the inscribed rectangle in the ellipse?

The parameter equation of the original elliptic equation is as follows
x=a·cosθ,y=b·sinθ
Where θ is the parameter, (substituting X and y, we can get (COS θ) ^ 2 + (sin θ) ^ 2 = 1, so it is true.)
The area of rectangle is s = 2acos θ· 2bsin θ
=2absin(2θ)
Because the maximum value of sin (2 θ) is 1
So the maximum area s of a rectangle is 2Ab