" (v) "2^(x)+2^(y)=12;x+y=5

Equation of parabola whose vertex is (2 5) and focus (2 2) is (A) (x-2)^(2)=12(y-5) (B) (x-2)^(2)=-12(y-5) (C) (x-2)^(2)=12(y-2) (D) (x-2)^(2)= -12(y-2)

The point to which the origin should be shifted in order to eliminate x and y terms in the equation 2x^(2) - 3y^(2) - 12x - 6y +5=0 is

Factorise : 3x^(5) y - 27 x^(4) y^(2) + 12x^(3) y^(3)

Divide : 36x^(2)y^(2) + 42 xy^(3) - 24 x^(3)y^(2) - 12 y^(5) by - 6y^(2)

If x^(2) + y^(2) + z^(2) + 12x + 4y + 5=0 find x^(12) + y+ z^(30) = ?

Equation of radical axis of the circles x^(2) + y^(2) - 3x - 4y + 5 = 0 and 2x^(2) + 2y^(2) - 10x - 12y + 12 = 0 is

Divide 36x^(2)y^(5)+42xy^(3)-24x^(3)y^(2)-12y^(5) by -6y^(2) .

The circles x^(2)+y^(2)+6x+6y=0 and x^(2)+y^(2)-12x-12y=0

If x,y,in R satisfies (x+5) ^(2) + (y-12) ^(2) = (14)^(2), then the minimum vlaue of sqrt (x ^(2) + y ^(2)) is "______"