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

If a hyperbola passes through the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, and the product of eccentricities is 1 , then

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

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

focus of hyperbola is $(5,0)$

focus of hyperbola is $(5 \sqrt{3}, 0)$

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