Question

find the equation of parabola with latus rectum joinin.g (2, 5), (2, -3).
find the equation of parabola with bertex in tje line y=6, axis parallel to y, latus rectum 6 and passing through (2, 8)

Answer 1

Recall that the parabola y^2 = 4px has
vertex at (0,0)
focus at (p,0)
directrix at x = -p
latus rectum of length 4p
You say the latus rectum has length=6, so I don't know what those two points are all about. They are clearly not the ends of the latus rectum. Especially since that line is vertical, yet you say the axis is vertical. I'll work a solution assuming a horizontal axis. You can revise it as you see fit.

So, let's say that the vertex is on the line y=6. That means we have
(y-6)^2 = 3/2 (x-h)
Since (2,8) is on the curve,
3/2 (2-h) = (8-6)^2
3 - 3/2 h = 4
h = 2/3
So, (y-6)^2 = 3/2 (x - 2/3)

Answer 2

#1.
Since the latus rectum is a vertical line, the parabola has a horizontal axis of
symmetry and takes the general form y^2 = 4ax, where 4a is the length of the
latus rectum.
4a = √((2-2)^2 + (5+3)^2) = 8
a = 2
Also we know that the focus is (2,1) , the midpoint of the latus rectum

so we know the vertex must be (0,1) since the the distance between the focus
and the vertex is a.
equation of parabola:
(y-1)^2 = 8x <------ expand it if need be

check:
is the point (2,5) on this?
LS = (y-1)^2 = 16
RS = 8(2) = 16, YES!

#2.
The latus rectum = 6, so 4a = 6
a = 3/2
Let the vertex be (p,6) and the equation is
(y-6) = 6(x-p)^2 **
but (2,8) lies on this, so
(8-6) = 6(2-p)^2
1/3 = (2-p)^2
2-p = ±1/√3 = ± √3/3
p = 2+√3/3 or p = 2-√3/3
= (6 + √3)/3 or (6 - √3)/3

sub p into ** and you will have 2 parabolas

check my arithmetic, I was expecting a "nicer" answer.