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1

### IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by

A
y $$-$$ x + 3 = 0
B
y + 3x $$-$$ 33 = 0
C
y + x $$-$$ 15 = 0
D
7 $$-$$ 2x + 12 = 0

## Explanation

The equation of normal is

y = mx $$-$$ 2m $$-$$ m3

As (9, 6) lies on it, 6 = 9m $$-$$ 2m $$-$$ m33 $$-$$ 7m + 6 = 0

The roots are m = 1, 2, $$-$$3. So the normal are

y = x $$-$$ 3, y = 2x $$-$$ 12, y = $$-$$3x + 33.

2

### IIT-JEE 2011 Paper 1 Offline

MCQ (More than One Correct Answer)
Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be reciprocal to that of the ellipse $${x^2} + 4{y^2} = 4$$. If the hyperbola passes through a focus of the ellipse, then
A
the equation of the hyperbola is $${{{x^2}} \over 3} - {{{y^2}} \over 2} = 1$$
B
a focus of the hyperbola is $$(2, 0)$$
C
theeccentricity of the hyperbola is $$\sqrt {{5 \over 3}}$$
D
The equation of the hyperbola is $${x^2} - 3{y^2} = 3$$

## Explanation

Ellipse is $${{{x^2}} \over {{2^2}}} + {{{y^2}} \over {{1^2}}} = 1$$.

$${1^2} = {2^2}(1 - {e^2}) \Rightarrow e = {{\sqrt 3 } \over 2}$$

Therefore, the eccentricity of the hyperbola is

$${2 \over {\sqrt 3 }} \Rightarrow {b^2} = {a^2}\left( {{4 \over 3} - 1} \right) \Rightarrow 3{b^2} = {a^2}$$

Foci of the ellipse are $$(\sqrt 3 ,\,0)$$ and $$( - \sqrt 3 ,\,0)$$.

Hyperbola passes through $$( \sqrt 3 ,\,0)$$.

$${3 \over {{a^2}}} = 1 \Rightarrow {a^2} = 3$$ and $${b^2} = 1$$

Therefore, the equation of hyperbola is $${x^2} - 3{y^2} = 3$$.

Focus of hyperbola is $$(ae,\,0) \equiv \left( {\sqrt 3 \times {2 \over {\sqrt 3 }},\,0} \right) \equiv (2,0)$$.

3

### IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $$A$$ and $$B$$ be two distinct points on the parabola $${y^2} = 4x$$. If the axis of the parabola touches a circle of radius $$r$$ having $$AB$$ as its diameter, then the slope of the line joining $$A$$ and $$B$$ can be
A
$$- {1 \over r}$$
B
$${1 \over r}$$
C
$${2 \over r}$$
D
$$- {2 \over r}$$

## Explanation

Let A $$\equiv$$ (t$$_1^2$$, 2t1) and B $$\equiv$$ (t$$_2^2$$, 2t2)

The centre of the circle = $$\left( {{{t_1^2 + t_2^2} \over 2},{t_1} + {t_2}} \right)$$

As the circle touches the x-axis thus $${t_1} + {t_2} = \pm \,r$$

Slope of $$AB = {2 \over {{t_1} + {t_2}}} = \pm \,{2 \over r}$$

4

### IIT-JEE 2009

MCQ (More than One Correct Answer)
An ellipse intersects the hyperbola $$2{x^2} - 2{y^2} = 1$$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes then
A
equation of ellipse is $${x^2} + 2{y^2} = 2$$
B
the foci of ellipse are $$\left( { \pm 1,0} \right)$$
C
equation of ellipse is $${x^2} + 2{y^2} = 4$$
D
the foci of ellipse are $$\left( { \pm \sqrt 2 ,0} \right)$$

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