1
IIT-JEE 2007
+3
-0.75
STATEMENT-1: The curve $$y = {{ - {x^2}} \over 2} + x + 1$$ is symmetric with respect to the line $$x=1$$. because

STATEMENT-2: A parabola is symmetric about its axis.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True.
2
IIT-JEE 2006
+3
-0.75
The axis of a parabola is along the line $$y = x$$ and the distances of its vertex and focus from origin are $$\sqrt 2$$ and $$2\sqrt 2$$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is
A
$${\left( {x + y} \right)^2} = \left( {x - y - 2} \right)$$
B
$${\left( {x - y} \right)^2} = \left( {x + y - 2} \right)$$
C
$${\left( {x - y} \right)^2} = 4\left( {x + y - 2} \right)$$
D
$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$
3
IIT-JEE 2006
+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
$$\left( A \right) - \left( r \right),\left( B \right) - \left( s \right),\left( C \right) - \left( q \right),\left( D \right) - \left( p \right)$$
4
IIT-JEE 2005 Screening
+2
-0.5
Tangent to the curve $$y = {x^2} + 6$$ at a point $$(1, 7)$$ touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$ at a point $$Q$$. Then the coordinates of $$Q$$ are
A
$$(-6, -11)$$
B
$$(-9, -13)$$
C
$$(-10, -15)$$
D
$$(-6, -7)$$
EXAM MAP
Medical
NEET