IIT-JEE 2010 Paper 1 Offline | Hyperbola Question 13 | Mathematics

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of the circle with $$AB$$ as its diameter is

$${x^2} + {y^2} - 12x + 24 = 0$$

$${x^2} + {y^2} + 12x + 24 = 0$$

$${x^2} + {y^2} + 24x - 12 = 0$$

$${x^2} + {y^2} - 24x - 12 = 0$$

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider a branch of the hyperbola $$${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$$$

with vertex at the point $$A$$. Let $$B$$ be one of the end points of its latus rectum. If $$C$$ is the focus of the hyperbola nearest to the point $$A$$, then the area of the triangle $$ABC$$ is

$$1 - \sqrt {{2 \over 3}} $$

$$\sqrt {{3 \over 2}} - 1$$

$$1 + \sqrt {{2 \over 3}} $$

$$\sqrt {{3 \over 2}} + 1$$

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

A hyperbola, having the transverse axis of length $$2\sin \theta ,$$ is confocal with the ellipse $$3{x^2} + 4{y^2} = 12.$$ Then its equation is

$${x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$$

$${x^2}{\sin ^2}\theta - {y^2}co{s^2}\theta = 1$$

$${x^2}{\cos ^2}\theta - {y^2}{\sin ^2}\theta = 1$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If the line $$62x + \sqrt 6 y = 2$$ touches the hyperbola $${x^2} - 2{y^2} = 4$$, then the point of contact is

$$\left( { - 2,\,\sqrt 6 } \right)$$

$$\left( { - 5,\,2\sqrt 6 } \right)$$

$$\left( {{1 \over 2},{1 \over {\sqrt 6 }}} \right)$$

$$\left( {4, - \,\sqrt 6 } \right)$$

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