Problem :
The vertex of the parabola is the point (a,b) and latus rectum is of length l. If the axis of the parabola is along the positive direction of y-axis, then its equation is
(a) $(x+a)^2= \frac{l}{2}(2y-2b)$
(b) $(x-a)^2 =\frac{l}{2}(2y-2b)$
(c) $(x+a)^2 =\frac{l}{4} (2y-2b)$
(d) $(x-a)^2 = \frac{l}{8}(2y-2b)$
My working :
Since the axis of parabola is along y axis therefore the equation of parabola will be of the type :
$x^2 = 4ay$
Now since length of latus rectum is 4a ( where a is the distance of focus from vertex) here it is $ 4a =l \Rightarrow a = \frac{l}{4}$
Now since vertex of parabola is shifted from origin (0,0) to (a,b) therefore the equation will become :
$(x-a)^2 = l (y-b) \Rightarrow (x-a)^2 = \frac{l}{2}(2y-2b)$ So answer is option b. but this is wrong answer option d is correct please guide on this.. thanks.