IIT-JEE 2010 Paper 1 Offline | Hyperbola Question 16 | Mathematics

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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 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 a common tangent with positive slope to the circle as well as to the hyperbola is

$$2x - \sqrt {5y} - 20 = 0$$

$$2x - \sqrt {5y} + 4 = 0$$

$$3x - 4y + 8 = 0$$

$$4x - 3y + 4 = 0$$

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 Paper 1 Offline

MCQ (Single Correct Answer)

-1

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

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

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

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

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

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