If the square of an ellipse x divided by the square of a plus the square of Y is equal to 1, and there is a point P on the ellipse so that its angle to two focal points E and F is 90 degrees, then the eccentricity of the ellipse is determined Range

If the square of an ellipse x divided by the square of a plus the square of Y is equal to 1, and there is a point P on the ellipse so that its angle to two focal points E and F is 90 degrees, then the eccentricity of the ellipse is determined Range

Knowledge point: the angle of the end point of the minor axis of the ellipse to the two focal points is the maximum angle of any point of the ellipse to the two focal points
In this problem, let B be an endpoint of the minor axis, then ∠ f1bf2 ≥ 90 °,
Thus B ≤ C, B & # 178; ≤ C & # 178;
a²-c²≤c²,a²≤2c²
√2/2≤e

X + 1 / 1 + X-2 / 2 = - x square - 8 / 4

This kind of expression is easy to lead to ambiguity, and you can add more brackets in the future
1/(x+1)+2/(x-2)=-8/(x^2-4)
[(x-2)+2(x+1)]/[(x+1)(x-2)]=-8/[(x-2)(x+2)]
3x/(x+1)=-8/(x+2)
3x(x+2)+8(x+1)=0
3x^2+14x+8=0
x=-4
x=-2/3

A ^ 2 of x plus B ^ 2 of Y minus the square of (a + b) of X + y
The answer is XY (x + y) of (ay BX) ^ 2. I want to know the process

a²/x+b²/y-(a+b)²/(x+y)=a²y(x+y)/xy(x+y)+b²x(x+y)/xy(x+y)-xy(a+b)²/xy(x+y)=[a²y(x+y)+b²x(x+y)-xy(a+b)²]/[xy(x+y)]=[a²xy+a²y²+b²x²+b...