1
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
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
$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = a$$
2
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
$$\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is :
A
$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$
B
$${1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$
C
$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$
D
$${1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$.
3
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
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
4
AIEEE 2003
MCQ (Single Correct Answer)
+4
-1
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
$${\left( {3x + 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}$$
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