The vertex of a parabola is (2, 2) and the coordinats of its two extremities of latus rectum are `(-2,0)` and (6, 0). Then find the equation of the parabola.

The vertex of the parabola y^2 + 4x = 0 is

Find the equation of a parabola having its focus at S(2,0) and one extremity of its latus rectum at (2, 2)

Consider the parabola y^(2) = 4ax (i) Write the equation of the rectum and obtain the x co-ordinates of the point of intersection of latus rectum and the parabola. (ii) Find the area of the parabola bounded by the latus rectum.

Find the vertex of the parabola x^2=2(2x+y)

Find the area of the parabola y^2=4ax and its latus rectum.

The vertex of a parabola is the point (a , b) and the latus rectum is of length ldot 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=l/2(2y-2b) (b) (x-a)^2=l/2(2y-2b) (c) (x+a)^2=l/4(2y-2b) (d) (x-a)^2=l/8(2y-2b)

Find the value of lambda if the equation (x-1)^2+(y-2)^2=lambda(x+y+3)^2 represents a parabola. Also, find its focus, the equation of its directrix, the equation of its axis, the coordinates of its vertex, the equation of its latus rectum, the length of the latus rectum, and the extremities of the latus rectum.

If the vertex of a parabola is the point (-3,0) and the directrix is the line x+5=0 , then find its equation.