A line is drawn form `A(-2,0)` to intersect the curve `y^2=4x` at `Pa n dQ` in the first quadrant such that `1/(A P)+1/(A Q)<1/4dot` Then the slope of the line is always. (a) `>sqrt(3)` (b) `<1/(sqrt(3))` (c) `>sqrt(2)` (d) `>1/(sqrt(3))`

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)

The length of the chord of the parabola y^2=x which is bisected at the point (2, 1) is 2sqrt(3) (b) 4sqrt(3) (c) 3sqrt(2) (d) 2sqrt(5)

On the line segment joining (1, 0) and (3, 0) , an equilateral triangle is drawn having its vertex in the fourth quadrant. Then the radical center of the circles described on its sides. (a) (3,-1/(sqrt(3))) (b) (3,-sqrt(3)) (c) (2,-1/sqrt(3)) (d) (2,-sqrt(3))

If the line y-sqrt(3)x+3=0 cut the parabola y^2=x+2 at P and Q , then A PdotA Q is equal to [where A=(sqrt(3),0)] (a) (2(sqrt(3)+2))/3 (b) (4sqrt(3))/2 (c) (4(2-sqrt(2)))/3 (d) (4(sqrt(3)+2))/3

Simplify: (1)/(sqrt2 +1) + (1)/(sqrt3 + sqrt2) + (1)/(sqrt4 + sqrt3)

If cosx=(2cosy-1)/(2-cosy), where x , y in (0,pi) then tan(x/2)xxcot(y/2) is equal to (a) sqrt(2) (b) sqrt(3) (c) 1/(sqrt(2)) (d) 1/(sqrt(3))

The area (in square units) bounded by the curves y=sqrt(x),2y-x+3=0 , x -axis and lying in the first quadrant is -

Tangent of acute angle between the curves y=|x^2-1| and y=sqrt(7-x^2) at their points of intersection is (a) (5sqrt(3))/2 (b) (3sqrt(5))/2 (5sqrt(3))/4 (d) (3sqrt(5))/4

If line x-2y-1=0 intersects parabola y^(2)=4x at P and Q, then find the point of intersection of normals at P and Q.

The area of the region is 1st quadrant bounded by the y-axis, y=(x)/(4), y=1 +sqrt(x), and y=(2)/(sqrt(x)) is