The ends of the latus rectum of the parabola `(x-2)^2=-6(y+1)` are

Find the latus rectum of the parabola (y-3)^2=6(x-2)

One end of latus rectum of the parabola (x+2)^(2)=-4(y+3) is

The ends of the latus rectum of the parabola x^(2)+10x-16y+25=0 are

The equation of the latus rectum of the parabola (y-2)^(2)=-4(x+2) is

The length of the latus rectum of the parabola (x+2) ^(2) = - 14 (y-5) is

The point of intersection of the tangents at the ends of the latus rectum of the parabola y^(2)=8 x is

The point of intersection of the two tangents at the ends of the latus rectum of the parabola (y+3)^(2)=8(x-2)

L and L are the ends of the latus rectum of the parabola x^(2)=6y. The equation of OL and OL where O is the origin is