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第1362题:共同渐近线的双曲线系



以下方程中 a>0,b>0a>0,b>0 哪两组双曲线有共同的渐近线? (  ).


A. x2a2y2b=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b}=1x2b2y2a=1\dfrac{x^2}{b^2}-\dfrac{y^2}{a}=1


B. x2a2y2b=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b}=1y2a2x2a=1\dfrac{y^2}{a^2}-\dfrac{x^2}{a}=1


C. x2a2y2b=1 \dfrac{x^2}{a^2}-\dfrac{y^2}{b}=1x2a2y2b=k\dfrac{x^2}{a^2}-\dfrac{y^2}{b}=k (k0)(k \ne 0)


D. x2(ka)2y2b=1\dfrac{x^2}{(ka)^2}-\dfrac{y^2}{b}=1x2a2y2b=k\dfrac{x^2}{a^2}-\dfrac{y^2}{b}=k  (k0,k1)(k \ne 0, k \ne 1)



共轭双曲线: 双曲线 x2a2y2b2=1 \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 的共轭双曲线是  x2a2+y2b2=1-\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 , 它们有共同的渐近线 y=±baxy= \pm \dfrac{b}{a}x  , 它们的离心率满足: 1e12+1e22=1\dfrac{1}{e_1^2}+\dfrac{1}{e_2^2}=1 .
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