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1

### IIT-JEE 2010 Paper 1 Offline

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

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

2

### IIT-JEE 2010 Paper 1 Offline

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. $$2x - \sqrt 5 y + 4 = 0$$

3

### IIT-JEE 2009

The locus of the orthocentre of the triangle formed by the lines $$\left( {1 + p} \right)x - py + p\left( {1 + p} \right) = 0,$$$$$\left( {1 + q} \right)x - qy + q\left( {1 + q} \right) = 0,$$$
and $$y=0$$, where $$p \ne q,$$ is
A
a hyperbola
B
a parabola
C
an ellipse
D
a straight line
4

### IIT-JEE 2009

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
A
$$\left( { \pm {{3\sqrt 5 } \over 2},\, \pm {2 \over 7}} \right)$$
B
$$\left( { \pm {{3\sqrt 5 } \over 2},\, \pm \sqrt {{{19} \over 4}} } \right)$$
C
$$\left( { \pm 2\sqrt 3 , \pm {1 \over 7}} \right)$$
D
$$\left( { \pm 2\sqrt 3 , \pm {{4\sqrt 3 } \over 7}} \right)$$

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