已知x,y满足x^2/4+y^2/b^2=1(b>1),求想x^2+2y的最大值
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发布时间:2024-10-07 06:27
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热心网友
时间:2024-11-26 19:25
答:
(x^2)/4+(y^2)/b^2=1
(y^2)/b^2=1-(x^2)/4<=1-0
y^2<=b^2
-b<=y<=b
椭圆(x^2)/a^2+(y^2)/b^2=1中一定有:
-a<=x<=a
-b<=y<=b
热心网友
时间:2024-11-26 19:26
由x²=4(1-y²/b²)大于等于0得到的,