The equation `|sqrt((x-2)^2+(y-1)^2)-sqrt((x+2)^2+y^2)|=c` will represent a hyperbola if (A) ` c in (0,6)` (B) ` c in (0,5)` (C) ` c in (0,sqrt17)` (D) ` c in (0,sqrt19)`

The equation | sqrt( (x- 2) ^(2) + (y-1) ^(2)) - sqrt( (x+ 2)^(2) +y^(2)) =c will represent a hyperbola if

The equation |sqrt(x^2+(y-1)^2)-sqrt(x^2+(y+1)^2)|=K will represent a hyperbola for (a) K in (0,2) (b) K in (-2,1) (c) K in (1,oo) (d) K in (0,oo)

The eccentricity of the hyperbola x ^(2) - y ^(2) = 4 is :a) sqrt3 b) 2 c) 1.5 d) sqrt2

Number of solution of the equation sqrt(x^(2))-sqrt((x-1)^(2))+sqrt((x-2)^(2))=sqrt(5), is (A) 0(B)1(C)2(D) More Than2

The eccentricity of the hyperbola 4x^(2) - y^(2) - 8x - 8y - 28 =0 is a)3 b) sqrt5 c) 2 d) sqrt7

If m_(1) and m_(2) are the roots of the equation x^(2) + ( sqrt(3) + 2) x + sqrt(3) - 1 =0 then prove that area of the triangle formed by the lines y=m_(1) x, y=m_(2) x and y=c is (c^2)/( 4) ( sqrt(33) + sqrt(11)) .

If the equation (a^2+b^2)x^2-2(a c+b d)x+c^2+d^2=0 has equal roots, then (a) a b=c d (b) a d=b c (c) a d=sqrt(b c) (d) a b=sqrt(c d)

An equilateral triangle whose two vertices are (-2, 0) and (2, 0) and which lies in the first and second quadrants only is circumscribed by a circle whose equation is : (A) sqrt(3)x^2 + sqrt(3)y^2 - 4x +4 sqrt(3)y = 0 (B) sqrt(3)x^2 + sqrt(3)y^2 - 4x - 4 sqrt(3)y = 0 (C) sqrt(3)x^2 + sqrt(3)y^2 - 4y + 4 sqrt(3)y = 0 (D) sqrt(3)x^2 + sqrt(3)y^2 - 4y - 4 sqrt(3) = 0

intf(x)dx=2(f(x))^3+C ,and f(0)=0 then f(x) is (A) x/2 (B) x^2/2 (C) sqrt(x/3) (D) 2 sqrt(x/3)

A circle C of radius 1 is inscribed in an equilateral triangle PQR . The points of contact of C with the sides PQ, QR, RP and D, E, F respectively. The line PQ is given by the equation sqrt(3) +y-6=0 and the point D is ((sqrt(3))/(2), 3/2) . Equation of the sides QR, RP are : (A) y=2/sqrt(3) x + 1, y = 2/sqrt(3) x -1 (B) y= 1/sqrt(3) x, y=0 (C) y= sqrt(3)/2 x + 1, y = sqrt(3)/2 x-1 (D) y=sqrt(3)x, y=0