JEE Advanced2011MathematicsHyperbolaActual
Let the eccentricity of the hyperbola x^2 a^2 - y^2 b^2 =1 be reciprocal to that of the ellipse x^2+4 y^2=4 . If the hyperbola passes through a focus of the ellipse, then
Options
- Athe equation of the hyperbola is x^2 3^2 - y^2 2^2 =1
- Ba focus of the hyperbola is (2,0)
- Cthe eccentricity of the hyperbola is 5 3
- Dthe equation of the hyperbola is x^2-3 y^2=3
Correct answer
B. a focus of the hyperbola is (2,0)
Step-by-step solution
Here, equation of ellipse aligned & x^2 4 + y^2 1 =1 & e^2=1- b^2 a^2 =1- 1 4 = 3 4 & e= 3 2 and focus ( a e, 0) & =( 3 , 0) & aligned For hyperbola x^2 a^2 - y^2 b^2 =1 , e₁^2=1+ b^2 a^2 where, e₁^2= 1 e^2 = 4 3 1+ b^2 a^2 = 4 3 b^2 a^2 = 1 3 and hyperbola passes through ( 3 , 0) . Now, 3 a^2 =1 a^2=3 From Eqs. (i) and (ii), we get b^2=1 Equation of hyperbola is x^2 3 - y^2 1 =1 Focus is ( a e, 0) . Now, ( 3 2 3 , 0 ) ( 2,0) Hence, both options (b) and (d) are correct.