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1

### AIEEE 2009

MCQ (Single Correct Answer)
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 $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$
2

### AIEEE 2008

MCQ (Single Correct Answer)
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 $${k^2} = 16 \Rightarrow k = \pm 4$$
3

### AIEEE 2007

MCQ (Single Correct Answer)
If one of the lines of $$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$ is a bisector of angle between the lines $$xy = 0,$$ then $$m$$ is
A
$$1$$
B
$$2$$
C
$$-1/2$$
D
$$-2$$

## Explanation

Equation of bisectors of lines, $$xy=0$$ are $$y = \pm x$$

$$\therefore$$ Put $$y = \pm \,x$$ in the given equation

$$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$$

$$\therefore$$ $$m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$$

$$\Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$$
4

### AIEEE 2007

MCQ (Single Correct Answer)
Let $$P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right)$$ and $$R = \left( {3,3\sqrt 3 } \right)$$ be three point. The equation of the bisector of the angle $$PQR$$ is
A
$${{\sqrt 3 } \over 2}x + y = 0$$
B
$$x + \sqrt {3y} = 0$$
C
$$\sqrt 3 x + y = 0$$
D
$$x + {{\sqrt 3 } \over 2}y = 0$$

## Explanation

Given : The coordinates of points $$P,Q,R$$ are $$(-1,0),$$

$$\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)$$ respectively

Slope of $$QR$$ $$= {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}$$

$$\Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}$$

$$\Rightarrow \angle RQX = {\pi \over 3}$$

$$\therefore$$ $$\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}$$

Let $$QM$$ bisects the $$\angle PQR,$$

$$\therefore$$ Slope of the line $$QM=tan$$ $${{2\pi } \over 3} = - \sqrt 3$$

$$\therefore$$ Equation of line $$QM$$ is $$\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$$

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

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