(y-k)^(2)=4a(x-h)

If the focus of the parabola (y-k)^(2)=4(x-h) always lies between the lines x+y=1 and x+y=3, then

End points of the Latusrectum of the parabola (x-h)^(2)=4a(y-k) is

If (x-h)^(2)=4a(y-k) represents a parabola then ends of latusrectum

Find the differential equation from the equation (x-h)^(2)+(y-k)^(2)=a^(2) by eliminating h and k .

If the focus of the parabola (y - k)^2 = 4(x - h) always lies between the lines x + y = 1 and x + y = 3, then

Suppose the parabola (y- k)^2= 4 (x - h), with vertex A, passes through O=(0,0) and L = (0,2). Let D be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at P. Then /_PDA is equal to

If two parabolas y^(2)-4a(x-k) and x^(2)=4a(y-k) have only one common point P, then the equation of normal to y^(2)=4a(x-k) at P is

If two parabolas y^(2)-4a(x-k) and x^(2)=4a(y-k) have only one common point P, then the equation of normal to y^(2)=4a(x-k) at P is

Find the vertex, focus and directix of the parabola (x-h)^(2)+4a(y-k)=0 .

If x = h + p Sec alpha, y = k + q cosec alpha then ((p)/(x-h))^(2) +((q)/(y-k))^(2)=