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1

AIEEE 2003

MCQ (Single Correct Answer)
If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then
A
$$pq = -1$$
B
$$p = q$$
C
$$p = -q$$
D
$$pq = 1$$.

Explanation

Equation of bisectors of second pair of straight lines is,

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

It must be identical to the first pair

$${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$$

from $$(1)$$ and $$(2)$$ $${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$$

$$\Rightarrow pq = - 1.$$
2

AIEEE 2003

MCQ (Single Correct Answer)
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 $$CA$$

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

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

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

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

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

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

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

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

$$y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.$$
3

AIEEE 2002

MCQ (Single Correct Answer)
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

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

$$\Rightarrow a = {{ - 3 \pm \sqrt {9 + 8} } \over 2} = {{ - 3 \pm \sqrt {17} } \over 2}$$
4

AIEEE 2002

MCQ (Single Correct Answer)
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 $$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$$

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

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