IIT-JEE 2005 Mains | Hyperbola Question 23 | Mathematics

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

IIT-JEE 1999

MCQ (Single Correct Answer)

-0.5

Let $$P$$ $$\left( {a\,\sec \,\theta ,\,\,b\,\tan \theta } \right)$$ and $$Q$$ $$\left( {a\,\sec \,\,\phi ,\,\,b\,\tan \,\phi } \right)$$, where $$\theta + \phi = \pi /2,$$, be two points on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$.

If $$(h, k)$$ is the point of intersection of the normals at $$P$$ and $$Q$$, then $$k$$ is equal to

$${{{a^2} + {b^2}} \over a}$$

$$ - \left( {{{{a^2} + {b^2}} \over a}} \right)$$

$${{{a^2} + {b^2}} \over b}$$

$$ - \left( {{{{a^2} + {b^2}} \over b}} \right)$$

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