1
IIT-JEE 2010 Paper 1 Offline
+4
-1
The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

A
$$2x - \sqrt {5y} - 20 = 0$$
B
$$2x - \sqrt {5y} + 4 = 0$$
C
$$3x - 4y + 8 = 0$$
D
$$4x - 3y + 4 = 0$$
2
IIT-JEE 2010 Paper 1 Offline
+4
-1
The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of the circle with $$AB$$ as its diameter is

A
$${x^2} + {y^2} - 12x + 24 = 0$$
B
$${x^2} + {y^2} + 12x + 24 = 0$$
C
$${x^2} + {y^2} + 24x - 12 = 0$$
D
$${x^2} + {y^2} - 24x - 12 = 0$$
3
IIT-JEE 2010 Paper 2 Offline
+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 $$\left( { - {9 \over 5},{8 \over 5}} \right)$$
4
IIT-JEE 2010 Paper 2 Offline
+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
$${x^2} + {y^2} - 2xy + 27x + 31y - 120 = 0$$
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