EllipseHard
Question
et the eccentricity of the hyperbola 
= 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then
= 1 be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, thenOptions
A.the equation of the hyperbola is 
B.a focus of the hyperbola is (2, 0)
C. the eccentricity of the hyperbola is 
D.the equation of the hyperbola is x2 - 3y2 = 3
Solution
Ellipse is 
12 = 22 (1 - e2) ⇒ e
∴ eccentricity of the hyperbola is
⇒ b2 = a2 
⇒ 3b2 = a2
Foci of the ellipse are (√3, 0) and (- √3, 0).
Hyperbola passes through (√3, 0)
= 1 ⇒ a2 = 3 and b2 = 1
∴ Equation of hyperbola is x2 - 3y2 = 3
Focus of hyperbola is (ae, 0)
(2, 0)
12 = 22 (1 - e2) ⇒ e
∴ eccentricity of the hyperbola is
⇒ b2 = a2 ⇒ 3b2 = a2 Foci of the ellipse are (√3, 0) and (- √3, 0).
Hyperbola passes through (√3, 0)
= 1 ⇒ a2 = 3 and b2 = 1∴ Equation of hyperbola is x2 - 3y2 = 3
Focus of hyperbola is (ae, 0)
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