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1

### 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 $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$
2

### 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.
3

### 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 $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$
4

### AIEEE 2008

The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is
A
1
B
2
C
-2
D
-4

## Explanation

Slope of $$PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$$

$$\therefore$$ Slope of perpendicular bisector of

$$PQ = \left( {k - 1} \right)$$

Also mid point of

$$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$$

Equation of perpendicular bisector is

$$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$$

$$\Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$$

$$\Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$$

$$\therefore$$ $$y$$-intercept $$= {{8 - {k^2}} \over { - 2}} = - 4$$

$$\Rightarrow$$ $$8 - {k^2} = - 8$$ or $${k^2} = 16 \Rightarrow k = \pm 4$$

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