IIT-JEE 2012 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

Tangents are drawn to the hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1,$$ parallel to the straight line $$2x - y = 1,$$ The points of contact of the tangents on the hyperbola are

$$\left( {{9 \over {2\sqrt 2 }},{1 \over {\sqrt 2 }}} \right)$$

$$\left( -{{9 \over {2\sqrt 2 }},-{1 \over {\sqrt 2 }}} \right)$$

$$\left( {3\sqrt 3 , - 2\sqrt 2 } \right)$$

$$\left( -{3\sqrt 3 , 2\sqrt 2 } \right)$$

IIT-JEE 2011 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be reciprocal to that of the ellipse $${x^2} + 4{y^2} = 4$$. If the hyperbola passes through a focus of the ellipse, then

the equation of the hyperbola is $${{{x^2}} \over 3} - {{{y^2}} \over 2} = 1$$

a focus of the hyperbola is $$(2, 0)$$

theeccentricity of the hyperbola is $$\sqrt {{5 \over 3}} $$

The equation of the hyperbola is $${x^2} - 3{y^2} = 3$$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1.25

Let a hyperbola passes through the focus of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the produced of eccentricities of given ellipse and hyperbola is $$1$$, then

the equation of hyperbola is $${{{x^2}} \over 9} + {{{y^2}} \over {16}} = 1$$

the equation of hyperbola is $${{{x^2}} \over 9} + {{{y^2}} \over {25}} = 1$$

focus of hyperbola is $$(5, 0)$$

vertex of hyperbola is $$\left( {5\sqrt 3 ,0} \right)$$

JEE Advanced Subjects