 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2010

The line $$L$$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $$L$$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $$L$$ and $$K$$ is
A
$$\sqrt {17}$$
B
$${{17} \over {\sqrt {15} }}$$
C
$${{23} \over {\sqrt {17} }}$$
D
$${{23} \over {\sqrt {15} }}$$

## Explanation

Slope of line $$L = - {b \over 5}$$

Slope of line $$K = - {3 \over c}$$

Line $$L$$ is parallel to line $$k.$$

$$\Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15$$

$$(13,32)$$ is a point on $$L.$$

$$\therefore$$ $${{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} = - {8 \over 5}$$

$$\Rightarrow b = - 20 \Rightarrow c = - {3 \over 4}$$

Equation of $$K:$$ $$y - 4x = 3$$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$\Rightarrow 4x - y + 3 = 0$$

Distance between $$L$$ and $$K$$

$$= {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$
2

### AIEEE 2009

Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point:
A
$$\left( {{5 \over 4},0} \right)$$
B
$$\left( {{5 \over 2},0} \right)$$
C
$$\left( {{5 \over 3},0} \right)$$
D
$$\left( {0,0} \right)$$

## Explanation

Given that

$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$

and $${{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}$$

$$\Rightarrow 3AP = AQ$$

$$\,\,\,\,\,\,$$ Let $$A = (x,y)$$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$

$$\Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + y{}^2$$

$$\Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2}$$

$$\Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0$$

$$\Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ A lies on the circle given by eq. $$(1).$$ As $$B$$ and $$C$$

also follow the same condition, - they must lie on the same circle.

$$\therefore$$ Center of circumcircle of $$\Delta ABC$$

$$=$$ Center of circle given by $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$
3

### AIEEE 2009

The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :
A
exactly one values of $$p$$
B
exactly two values of $$p$$
C
more than two values of $$p$$
D
no value of $$p$$

## Explanation

If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$

and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$

are perpendicular to a common line then these lines -

must be parallel to each other,

$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}$$

$$\Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0$$

$$\Rightarrow p = - 1$$

$$\therefore$$ $$p$$ can have exactly one value.
4

### AIEEE 2009

The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :
A
$${{2\sqrt 3 } \over 8}$$
B
$${{3\sqrt 2 } \over 5}$$
C
$${{\sqrt 3 } \over 4}$$
D
$${{3\sqrt 2 } \over 8}$$

## Explanation

Let $$\left( {{a^2},a} \right)$$ be the point of shortest distance on $$x = {y^2}$$

Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$

is given by

$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$

It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$

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