有结论 $\dfrac{\sin^2\theta}{{e_1}^2} + \dfrac{\cos^2\theta}{{e_2}^2} = 1$ 进行推广 $\dfrac{\sin^2\alpha}{{e_1}^2} + \dfrac{\cos^2\beta}{{e_2}^2} = \sin^2\alpha + \cos^2\beta$ 则 $S=b…