A line is drawn form `A(-2,0)` to intersect the curve `y^2=4x` at `P` and `Q` in the first quadrant such that `1/(A P)+1/(A Q) lt 1/4` Then the slope of the line is always. (A) `gt sqrt(3)` (B) `lt 1/(sqrt(3))` (C) `gt sqrt(2)` (D) `gt 1/(sqrt(3))`

Find the sum of the first n terms of the series 1/(sqrt(2) + sqrt(3)) + 1/(sqrt(3) + sqrt(4)) +...

A line ax +by +c = 0 through the point A(-2,0) intersects the curve y^(2)=4a in P and Q such that (1)/(AP) +(1)/(AQ) =(1)/(4) (P,Q are in 1st quadrant). The value of sqrt(a^(2)+b^(2)+c^(2)) is

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.

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

P is a point on the line y+2x=1, and Q and R two points on the line 3y+6x=6 such that triangle P Q R is an equilateral triangle. The length of the side of the triangle is (a) 2/(sqrt(5)) (b) 3/(sqrt(5)) (c) 4/(sqrt(5)) (d) none of these

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))

Prove that pi/6 lt int_0^1(dx) /(sqrt(4-x^2-x^3)) lt pi/(4sqrt(2))

P and Q are points on the line joining A(-2,5) and B(3,1) such that A P=P Q=Q B . Then, the distance of the midpoint of P Q from the origin is (a) 3 (b) (sqrt(37))/2 (c) 4 (d) 3.5

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)

If two different tangents of y^2=4x are the normals to x^2=4b y , then (a) |b|>1/(2sqrt(2)) (b) |b| 1/(sqrt(2)) (d) |b|<1/(sqrt(2))