The vertex of a parabola is the point `(a , b)` and the latus rectum is of length `l`. If the axis of the parabola is parallel to the y-axis and the parabola is concave upward, then its equation is a. `(x+a)^2=1/2(2y-2b)` b. `(x-a)^2=1/2(2y-2b)` c. `(x+a)^2=1/4(2y-2b)` d. `(x-a)^2=1/8(2y-2b)`

The vertex of the parabola y^(2) =(a-b)(x -a) is

Find the length of the latus rectum of the parabola x^(2) = -8y .

Find the equation of parabola whose Vertex is (-1,-2) latus rectum is 4 and axis is parallel to y axis

Find the vertex, focus, directrix and length of the latus rectum of the parabola y^2 - 4y-2x-8=0

Find the vertex and length of latus rectum of the parabola x^(2)=-4(y-a) .

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 (a) (x+y)^2=(x-y-2) (b) (x-y)^2=(x+y-2) (c) (x-y)^2=4(x+y-2) (d) (x-y)^2=8(x+y-2)

x^(2)=6y . Find latus rectum and axis of parabola

If (a ,b) is the midpoint of a chord passing through the vertex of the parabola y^2=4(x+1), then prove that 2(a+1)=b^2

I : The length of the latus rectum of the parabola y^(2)+8x-2y+17=0 is 8 . II: The focal distance of the point (9,6) on the parabola y^(2)=4x is 10

Find the vertex, focus, axis , latus rectum and directrix of the parabola y^(2)+4x+6y+17=0