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)`

7(x-2y)^(2)-25(x-2y)+12

7(x-2y)^2-25(x-2y)+12

If the vertex of the parabola y=3x^(2)-12x+9 is (a b) then parabola whose vertex is (b a) is (are): (A) y=x^(2)+6x+11 (B) y=x^(2)-7x+3 (C) y=-2x^(2)-12x-16 (D) y=-2x^(2)+16x-13

If the vertex of the parabola y=3x^(2)-12x+9 is (a b) then parabola whose vertex is (b a) is (are): (A) y=x^(2)+6x+11 (B) y=x^(2)-7x+3 (C) y=-2x^(2)-12x-16 (D) y=-2x^(2)+16x-13

If the vertex of the parabola y=3x^(2)-12x+9 is (a b) then parabola whose vertex is (b a) is (are): (A) y=x^(2)+6x+11 (B) y=x^(2)-7x+3 (C) y=-2x^(2)-12x-16 (D) y=-2x^(2)+16x-13 205067267

If the vertex of the parabola y=3x^(2)-12x+9 is (a b) then parabola whose vertex is (b a) is (are): (A) y=x^(2)+6x+11 (B) y=x^(2)-7x+3 (C) y=-2x^(2)-12x-16 (D) y=-2x^(2)+16x-13

The axis of a parabola is along the line y=x and the distance of its vertex and focus from the origin are sqrt(2) and 2sqrt(2), respectively.If vertex and focus both lie in the first quadrant, then the equation of the parabola is (x+y)^(2)=(x-y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=8(x+y-2)

(1) (5)/(2x)+(2)/(3y)=7,(3)/(x)+(2)/(y)=12

The vertex of the parabola (y-2)^(2)=16(x-1) is (1,2) b.(-1,2)c(1,-2)d.(2,1)

Equation of the parabola whose focus is centre of the circle " x^(2)+y^(2)=4x+6y " and whose directrix is the line " x-y=0 " is A) " x^(2)+y^(2)+2xy-8x-12y+26=0 B ) " x^(2)+y^(2)-2xy-8x-12y-26=0 C) " x^(2)+y^(2)+2xy+8x+12y+26=0 D) " x^(2)+y^(2)-2xy-8x-12y+26=0