1

### IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)
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
$$\left( {{8 \over 25},{{7} \over 5}} \right)$$
2

### IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

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
$${x^2} + {y^2} - 2xy + 27x + 31y - 120 = 0$$
3

### IIT-JEE 2010 Paper 2 Offline

MCQ (Single Correct Answer)

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 1 Offline

MCQ (Single Correct Answer)
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
$${x^2} + {y^2} - 24x - 12 = 0$$

## Explanation

A point on hyperbola is (3sec$$\theta$$, 2tan$$\theta$$).

It lies on the circle, so

$$9{\sec ^2}\theta + 4{\tan ^2}\theta - 24\sec \theta = 0$$.

$$\Rightarrow 13{\sec ^2}\theta - 24\sec \theta - 4 = 0 \Rightarrow \sec \theta = 2, - {2 \over {13}}$$

Therefore, $$\sec \theta = 2 \Rightarrow \tan \theta = \sqrt 3$$

The point of intersection are $$A(6,2\sqrt 3 )$$ and $$B(6, - 2\sqrt 3 )$$

Hence, The circle with AB as diameter is

$${(x - 6)^2} + {y^2} = {(2\sqrt 3 )^2} \Rightarrow {x^2} + {y^2} - 12x + 24 = 0$$

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