若实数x,y满足x2+4y2=4x,求x2-y2的最大值和最小值

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x2+4y2=4x
x²-4x+4y²=0
(x-2)²+4y²=4
0≤x≤4-1≤y≤1
x2+4y2=4x得y²=(4x-x²)/4
x²-y²
=x²-(4x-x²)/4
=5x²/4-x
=(5/4)(x²-4x/5)
=(5/4)(x-2/5)²-1/5
当x=2/5时,x²-y²有最小值为-1/5
当x=4时,x²-y²有最大值为16
======以下答案可供参考======
供参考答案1:
x^2-4x+4y^2=0
(x-2)^2+4y^2=4
(x-2)^2/4+y^2=1
所以设x=2cosa+2 y=sina (0x^2-y^2
=4cos^2a+8cosa+4-sin^2a
=5cos^2a+8cosa+3
=5(cosa+4/5)^2-1/5
所以最大值为16，最小值为-1/5