Given that C is the half focal length of the ellipse x2a2 + y2b2 = 1 (a > b > 0), then the value range of B + Ca is______ ．

Given that C is the half focal length of the ellipse x2a2 + y2b2 = 1 (a > b > 0), then the value range of B + Ca is______ ．

According to the meaning of the question, (B + Ca) 2 = B2 + C2 + 2bca2 = B2 + C2 + 2bcb2 + C2 = 1 + 2bcb2 + C2 ≤ 2, i.e. 1 ＜ (B + Ca) 2 ≤ 2, the solution is 1 ＜ B + Ca ≤ 2, so the answer is (1,2]

Given x2 + y2 = 1, what is the range of Y / (x + 2)?

Let Y / (x + 2) = K
So y = K (x + 2)
Substitute the above formula into x2 + y2 = 1 to get
x2+k2(x2+4x+4)=1
Well organized
（1+k2）x2+4k2x + 4k2-1=0
Triangle = 16k4-4 (1 + K2) (4k2-1) > = 0
The solution is (- radical 3 / 3, radical 3 / 3)

4x2-y2/x2-2xy+y2 +(2x-y)/2x2-2xy

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