IIT-JEE 2007 Paper 1 Offline | Hyperbola Question 20 | Mathematics

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

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

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

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

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

IIT-JEE 2005 Mains

MCQ (Single Correct Answer)

-1

Tangents are drawn from any point on the hyperbola $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$ to the circle $$x^{2}+y^{2}=9$$. Find the locus of mid-point of the chord of contact.

$${{{x^2}} \over 4} + {{{y^2}} \over 9} = {{{{({x^2} + {y^2})}^2}} \over {81}}$$

$${{{x^2}} \over 4} - {{{y^2}} \over 9} = {{{{({x^2} + {y^2})}^2}} \over {81}}$$

$${{{x^2}} \over 9} + {{{y^2}} \over 4} = {{{{({x^2} + {y^2})}^2}} \over {81}}$$

$${{{x^2}} \over 9} - {{{y^2}} \over 4} = {{{{({x^2} + {y^2})}^2}} \over {81}}$$

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

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

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

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

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

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.5

For hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ which of the following remains constant with change in $$'\alpha '$$

abscissae of vertices

abscissae of foci

eccentricity

directrix

PREVIOUS NEXT

JEE Advanced Subjects

Browse all chapters by subject