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b^2x^2+a^2y^2 = a^b^2, 两边对x求导，得
2b^2x+2a^2yy' = 0, y' = -b^2x/(a^2y) = -(b^2/a^2)(x/y)
y'' = -(b^2/a^2)(y-xy')/y^2 = -(b^2/a^2)[y+(b^2/a^2)(x^2/y)]/y^2
= -(b^2/a^2)[a^2y^2+b^2x^2]/(a^2y^3)
= -(b^2/a^2)(a^2b^2)/(a^2y^3) = -b^4/(a^2y^3)
这是个椭圆方程？
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