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

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

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 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
$$4x - 3y + 4 = 0$$

Explanation

A tangent to $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ is

$$y = mx + \sqrt {9{m^2} - 4} ,\,m > 0$$ .... (1)

A tangent to $${(x - 4)^2} + {y^2} = 16$$ is

$$xy = m(x - 4) + 4\sqrt {1 + {m^2}} $$ ..... (2)

Comparing (1) and (2),

$$\sqrt {9{m^2} - 4} = - 4m + 4\sqrt {1 + {m^2}} \Rightarrow \sqrt {9 - {4 \over {{m^2}}}} = - 4 + 4\sqrt {1 + {1 \over {{m^2}}}} $$

Let $${1 \over {{m^2}}} = t$$, we have $$\sqrt {9 - 4t} = - 4 + 4\sqrt {1 + t} $$

Squaring, we have

$$ \Rightarrow 9 - 4t = 16 + 16(1 + t) - 32\sqrt {1 + t} \Rightarrow 32\sqrt {1 + t} = 23 + 20t$$

Again squaring $$1024(1 + t) = 529 + 920t + 400{t^2}$$

$$ \Rightarrow 400{t^2} - 104t - 495 = 0 \Rightarrow t = {5 \over 4}$$

Thus $${m^2} = {4 \over 5},\,m = {2 \over {\sqrt 5 }}$$

The tangent is $$y = {2 \over {\sqrt 5 }}x + {4 \over {\sqrt 5 }}$$ i.e. $$2x - \sqrt 5 y + 4 = 0$$

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