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1

### AIEEE 2003

A square of side a lies above the $$x$$-axis and has one vertex at the origin. The side passing through the origin makes an angle $$\alpha \left( {0 < \alpha < {\pi \over 4}} \right)$$ with the positive direction of x-axis. The equation of its diagonal not passing through the origin is
A
$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$
B
$$y\left( {\cos \alpha - \sin \alpha } \right) - x\left( {\sin \alpha - \cos \alpha } \right) = a$$
C
$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha - \cos \alpha } \right) = a$$
D
$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = a$$

## Explanation

Co-ordinate of $$A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)$$

Equation of $$OB,$$

$$y = \tan \left( {{\pi \over 4} + \alpha } \right)x$$

$$CA{ \bot ^r}$$ to $$OB$$

$$\therefore$$ slope of $$CA=-$$ $$\cot \left( {{\pi \over 4} + \alpha } \right)$$

Equation of $$CA$$

$$y - a\sin \alpha = - cot\left( {{\pi \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)$$

$$\Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right)$$

$$\Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha - x} \right)$$

$$\Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {1 - \tan \alpha } \right)$$

$$\Rightarrow \left( {y - a\sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {\cos \alpha - \sin \alpha } \right)$$

$$\Rightarrow y\left( {\cos + \sin \alpha } \right) - a\sin \alpha \cos \alpha - a{\sin ^2}\alpha$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha - a\cos \alpha \sin \alpha - x\left( {\cos \alpha - \sin \alpha } \right)$$

$$\Rightarrow y\left( {\cos \alpha + sin\alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$

$$y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.$$
2

### AIEEE 2002

The pair of lines represented by $$3a{x^2} + 5xy + \left( {{a^2} - 2} \right){y^2} = 0$$\$

are perpendicular to each other for
A
two values of $$a$$
B
$$\forall \,a$$
C
for one value of $$a$$
D
for no values of $$a$$

## Explanation

$$3a + {a^2} - 2 = 0 \Rightarrow {a^2} + 3a - 2 = 0;$$

$$\Rightarrow a = {{ - 3 \pm \sqrt {9 + 8} } \over 2} = {{ - 3 \pm \sqrt {17} } \over 2}$$
3

### AIEEE 2002

If the pair of lines

$$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$

intersect on the $$y$$-axis then
A
$$2fgh = b{g^2} + c{h^2}$$
B
$$b{g^2} \ne c{h^2}$$
C
$$abc = 2fgh$$
D
none of these

## Explanation

Put $$x=0$$ in the given equation

$$\Rightarrow b{y^2} + 2fy + c = 0.$$

For unique point of intersection $${f^2} - bc = 0$$

$$\Rightarrow a{f^2} - abc = 0.$$

Since $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$

$$\Rightarrow 2fgh - b{g^2} - c{h^2} = 0$$
4

### AIEEE 2002

Locus of mid point of the portion between the axes of

$$x$$ $$cos$$ $$\alpha + y\,\sin \alpha = p$$ where $$p$$ is constant is
A
$${x^2} + {y^2} = {4 \over {{p^2}}}$$
B
$${x^2} + {y^2} = 4{p^2}$$
C
$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {2 \over {{p^2}}}$$
D
$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}$$

## Explanation

Equation of $$AB$$ is

$$x\cos \alpha + y\sin \alpha = p;$$

$$\Rightarrow {{x\cos \alpha } \over p} + {{y\sin \alpha } \over p} = 1;$$

$$\Rightarrow {x \over {p/\cos \alpha }} + {y \over {p/\sin \alpha }} = 1$$

So co-ordinates of $$A$$ and $$B$$ are

$$\left( {{p \over {\cos \alpha }},0} \right)$$ and $$\left( {0,{p \over {\sin \alpha }}} \right);$$

So coordinates of midpoint of $$AB$$ are

$$\left( {{p \over {2\cos \,\alpha }},{p \over {2\sin \alpha }}} \right) = \left( {{x_1},{y_1}} \right)\left( {let} \right);$$

$${x_1} = {p \over {2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = {p \over {2\sin \alpha }};$$

$$\Rightarrow \cos \alpha = p/2{x_1}$$ and $$\sin \alpha = p/2{y_1};$$

$${\cos ^2}\alpha + {\sin ^2}\alpha = 1 \Rightarrow {{{p^2}} \over 4}\left( {{1 \over {{x_1}^2}} + {1 \over {{y_1}^2}}} \right) = 1$$

Locus of $$\left( {{x_1},{y_1}} \right)$$ is $${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.$$

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