IIT-JEE 2007 | Hyperbola Question 17 | Mathematics

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

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

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

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

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

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

IIT-JEE 1999

MCQ (Single Correct Answer)

-0.5

If $$x$$ $$=$$ $$9$$ is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$ then the equation of the vcorresponding pair of tangents is

$$9{x^2} - 8{y^2} + 18x - 9 = 0$$

$$9{x^2} - 8{y^2} - 18x + 9 = 0$$

$$9{x^2} - 8{y^2} - 18x - 9 = 0$$

$$9{x^2} - 8{y^2} + 18x + 9 = 0$$

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