1
IIT-JEE 2010 Paper 2 Offline
Numerical
+4
-0
Let $${a_1},\,{a_{2\,}},\,{a_3}$$......,$${a_{11}}$$ be real numbers satisfying $${a_1} = 15,27 - 2{a_2} > 0\,\,and\,\,{a_k} = 2{a_{k - 1}} - {a_{k - 2}}\,\,for\,k = 3,4,........11$$. if $$\,\,\,{{a_1^2 + a_2^2 + a_{11}^2} \over {11}} = 90$$, then the value of $${{{a_1} + {a_2} + .... + {a_{11}}} \over {11}}$$ is equal to
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2
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-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

A
$$(3,0)$$ and $$(0,2)$$
B
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
C
$$\left( { - {8 \over 5},{{2\sqrt {161} } \over {15}}} \right)$$ and $$(0,2)$$
D
$$(3,0)$$ and $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
3
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-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

A
$$9{x^2} + {y^2} - 6xy - 54x - 62y + 241 = 0$$
B
$${x^2} + 9{y^2} + 6xy - 54x + 62y - 241 = 0$$
C
$$9{x^2} + 9{y^2} - 6xy - 54x - 62y - 241 = 0$$
D
$${x^2} + {y^2} - 2xy + 27x + 31y - 120 = 0$$
4
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-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

A
$$\left( {5,{8 \over 7}} \right)$$
B
$$\left( {{7 \over 5},{{25} \over 8}} \right)$$
C
$$\left( {{11 \over 5},{{8} \over 5}} \right)$$
D
$$\left( {{8 \over 25},{{7} \over 5}} \right)$$
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