IIT-JEE 2010 Paper 2 Offline | Conic Sections Question 55 | Mathematics

IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Tangents are drawn from the point $$P(3, 4)$$ to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$$ touching the ellipse at points $$A$$ and $$B$$.

The equation of the locus of the point whose distances from the point $$P$$ and the line $$AB$$ are equal, is

$$9{x^2} + {y^2} - 6xy - 54x - 62y + 241 = 0$$

$${x^2} + 9{y^2} + 6xy - 54x + 62y - 241 = 0$$

$$9{x^2} + 9{y^2} - 6xy - 54x - 62y - 241 = 0$$

$${x^2} + {y^2} - 2xy + 27x + 31y - 120 = 0$$

IIT-JEE 2009 Paper 2 Offline

MCQ (Single Correct Answer)

-1

The normal at a point $$P$$ on the ellipse $${x^2} + 4{y^2} = 16$$ meets the $$x$$- axis $$Q$$. If $$M$$ is the mid point of the line segment $$PQ$$, then the locus of $$M$$ intersects the latus rectums of the given ellipse at the points

$$\left( { \pm {{3\sqrt 5 } \over 2},\, \pm {2 \over 7}} \right)$$

$$\left( { \pm {{3\sqrt 5 } \over 2},\, \pm \sqrt {{{19} \over 4}} } \right)$$

$$\left( { \pm 2\sqrt 3 , \pm {1 \over 7}} \right)$$

$$\left( { \pm 2\sqrt 3 , \pm {{4\sqrt 3 } \over 7}} \right)$$

IIT-JEE 2009 Paper 2 Offline

MCQ (Single Correct Answer)

-1

The locus of the orthocentre of the triangle formed by the lines

$$(1 + p)x - py + p(1 + p) = 0, $$

$$(1 + q)x - qy + q(1 + q) = 0$$

and $$y = 0$$, where $$p \ne q$$, is :

a hyperbola.

a parabola.

an ellipse.

a straight line.

IIT-JEE 2009 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The line passing through the extremity $$A$$ of the major axis and extremity $$B$$ of the minor axis of the ellipse $${x^2} + 9{y^2} = 9$$ meets its auxiliary circle at the point $$M$$. Then the area of the triangle with vertices at $$A$$, $$M$$ and the origin $$O$$ is

$${{31} \over {10}}$$

$${{29} \over {10}}$$

$${{21} \over {10}}$$

$${{27} \over {10}}$$

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