IIT-JEE 2010 Paper 2 Offline | Ellipse Question 15 | 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 coordinates of $$A$$ and $$B$$ are

$$(3,0)$$ and $$(0,2)$$

$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$

$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$(0,2)$$

$$(3,0)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$

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 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 orthocentre of the triangle $$PAB$$ is

$$\left( {5,{8 \over 7}} \right)$$

$$\left( {{7 \over 5},{{25} \over 8}} \right)$$

$$\left( {{11 \over 5},{{8} \over 5}} \right)$$

$$\left( {{8 \over 25},{{7} \over 5}} \right)$$

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

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

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

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

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

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