MathematicsHard74×since 2004Q3767Let x2a2+y2b2=1,a>b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}a2x2+b2y2=1,a>b be an ellipse, whose eccentricity is 12\frac{1}{\sqrt{2}}21 and the length of the latusrectum is 14\sqrt{14}14. Then the square of the eccentricity of x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2−b2y2=1 is :A3B72{7 \over 2}27C32{3 \over 2}23D52{5 \over 2}25Check answerSkip