1
IIT-JEE 2006
MCQ (More than One Correct Answer)
+5
-1.25
Let a hyperbola passes through the focus of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the produced of eccentricities of given ellipse and hyperbola is $$1$$, then
A
the equation of hyperbola is $${{{x^2}} \over 9} + {{{y^2}} \over {16}} = 1$$
B
the equation of hyperbola is $${{{x^2}} \over 9} + {{{y^2}} \over {25}} = 1$$
C
focus of hyperbola is $$(5, 0)$$
D
vertex of hyperbola is $$\left( {5\sqrt 3 ,0} \right)$$
2
IIT-JEE 2006
MCQ (Single Correct Answer)
+6
-1.5
Match the following : $$(3, 0)$$ is the pt. from which three normals are drawn to the parabola $${y^2} = 4x$$ which meet the parabola in the points $$P, Q $$ and $$R$$. Then

Column $${\rm I}$$
(A) Area of $$\Delta PQR$$
(B) Radius of circumcircle of $$\Delta PQR$$
(C) Centroid of $$\Delta PQR$$
(D) Circumcentre of $$\Delta PQR$$

Column $${\rm I}$$$${\rm I}$$
(p) $$2$$
(q) $$5/2$$
(r) $$(5/2, 0)$$
(s) $$(2/3, 0)$$

A
$$\left( A \right) - \left( p \right),\left( B \right) - \left( q \right),\left( C \right) - \left( s \right),\left( D \right) - \left( r \right)$$
B
$$\left( A \right) - \left( p \right),\left( B \right) - \left( q \right),\left( C \right) - \left( r \right),\left( D \right) - \left( s \right)$$
C
$$\left( A \right) - \left( s \right),\left( B \right) - \left( r \right),\left( C \right) - \left( p \right),\left( D \right) - \left( q \right)$$
D
$$\left( A \right) - \left( r \right),\left( B \right) - \left( s \right),\left( C \right) - \left( q \right),\left( D \right) - \left( p \right)$$
3
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-0.75
One angle of an isosceles $$\Delta $$ is $${120^ \circ }$$ and radius of its incircle $$ = \sqrt 3 $$. Then the area of the triangle in sq. units is
A
$$7 + 12\sqrt 3 $$
B
$$12 - 7\sqrt 3 $$
C
$$12 + 7\sqrt 3 $$
D
$$4\pi $$
4
IIT-JEE 2006
MCQ (More than One Correct Answer)
+5
-1.25
In $$\Delta ABC$$, internal angle bisector of $$\angle A$$ meets side $$BC$$ in $$D$$. $$DE \bot AD$$ meets $$AC$$ in $$E$$ and $$AB$$ in $$F$$. Then
A
$$AE$$ is $$HM$$ of $$b$$ and $$c$$
B
$$AD$$ $$ = {{2bc} \over {b + c}}\cos {A \over 2}$$
C
$$EF$$ $$ = {{4bc} \over {b + c}}\sin {A \over 2}$$
D
$$\Delta AEF$$ is isosceles
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