Let the eccentricity of the hyperbola be reciprocal to that of the ellipse

MATHEMATICS · Single correct · CBSE · Mathematics · Conic Sections · Conic Sections

Problem

Let the eccentricity of the hyperbola x2a2−y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 be reciprocal to that of the ellipse x2+4y2=4x^2 + 4y^2 = 4 . If the hyperbola passes through a focus of the ellipse, then (1) the equation of the hyperbola is x23−y22=1\frac{x^2}{3} - \frac{y^2}{2} = 1 (2) a focus of the hyperbola is (2,0)(2, 0) (3) the eccentricity of the hyperbola is 53\sqrt{\frac{5}{3}} (4) the equation of the hyperbola is x2−3y2=3x^2 - 3y^2 = 3

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