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AIEEE 2003

MCQ (Single Correct Answer)
If $${x_1},{x_2},{x_3}$$ and $${y_1},{y_2},{y_3}$$ are both in G.P. with the same common ratio, then the points $$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$
A
are vertices of a triangle
B
lie on a straight line
C
lie on an ellipse
D
lie on a circle

Explanation

Taking co-ordinates as

$$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$$

Then slope of line joining

$$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$$

and slope of line joining $$(x,y)$$ and $$(xr, yr)$$

$$ = {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}$$

$$\therefore$$ $${m_1} = {m_2}$$

$$ \Rightarrow $$ Points lie on the straight line.
2

AIEEE 2003

MCQ (Single Correct Answer)
Locus of centroid of the triangle whose vertices are $$\left( {a\cos t,a\sin t} \right),\left( {b\sin t, - b\cos t} \right)$$ and $$\left( {1,0} \right),$$ where $$t$$ is a parameter, is
A
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
B
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} - {b^2}$$
C
$${\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
D
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$

Explanation

$$x = {{a\cos t + b\sin t + 1} \over 3}$$

$$ \Rightarrow a\cos t + b\sin t = 3x - 1$$

$$y = {{a\sin t - b\cos t} \over 3}$$

$$ \Rightarrow a\sin t - b\cos t = 3y$$
3

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

$$ \Rightarrow pq = - 1.$$
4

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 $$CA$$

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

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

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

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

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

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

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

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

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

Questions Asked from Straight Lines and Pair of Straight Lines

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