find the equation of a parabola whose vertex is on the y axis, latus rectum 5, passing through (-3/2, -16/5) and whose axis on the y-axis. The answers are x^2 = 5(y+ 73/20) and x^2 = -5(y+11/4). HELP ME PLEASE. Show it to me step by step.

Question

find the equation of a parabola whose vertex is on the y axis, latus rectum 5, passing through (-3/2, -16/5) and whose axis on the y-axis.  The answers are x^2 = 5(y+ 73/20) and x^2 = -5(y+11/4).  HELP ME PLEASE. Show it to me step by step.

anonymous · Answer

kk we can do this

anonymous · Answer

we know the $x$ coordinate of the vertex is $0$ because it is on the $y$ axis

anonymous · Answer

and we know $4p=5$ or $4p=-5$ because of that "latus rectum" bit

anonymous · Answer

yes yes

anonymous · Answer

since it has axis of symmetry as the $y$ axis you know the $x$ part is squared

anonymous · Answer

so it looks like 
$$x^2=5(y-k)$$ or 
$$x^2=-5(y-k)$$

anonymous · Answer

put $x=-\frac{3}{2}$ and $y=-\frac{16}{5}$ and solve for $k$

anonymous · Answer

i am going to go to bed soon, and try to get this "rectum" business out of my head

anonymous · Answer

yes! Thank you so much :)) Goodnight!

anonymous · Answer

wait. how bout 73/20? the other one?

anonymous · Answer

okay got it now

anonymous · Answer

you have two equations to solve  for $k$
one with $5$ and the other with $-5$
i didn't do it, but i imagine that is how you are going to get two answers

anonymous · Answer

thank you :)))

anonymous · Answer

yw