Question 1148688
<br>
Put the equation in vertex form,<br>
{{{(x-h) = (1/(4p))(y-k)^2}}}
or
{{{x = (1/(4p))(y-k)^2+h}}}<br>
When you have done that, p is the directed distance from the directrix to the vertex, and it is the directed distance from the vertex to the focus.  Finally, 4p is the length of the latus rectum.<br>
{{{y^2-8x-4y-20 = 0}}}  given form
{{{y^2-4y = 8x+20}}}  isolate the y terms
{{{y^2-4y+4 = 8x+24}}}  complete the square in y
{{{(y-2)^2 = 8(x+3)}}}  put in vertex form
{{{(x+3) = (1/8)(y-2)^2}}}  done. vertex (-3,2); 4p=8<br>
The length of the latus rectum is 8.<br>
In more detail....<br>
The vertex is (-3,2)<br>
4p=8, so p=2.<br>
Since the y term is squared and p is positive, the parabola opens to the right.<br>
p=2 is the directed distance from the directrix to the vertex, so the directrix is 2 units to the left of the vertex, at x=-5.<br>
p=2 is also the directed distance from the vertex to the focus, so the focus is 2 units to the right of the vertex, at (-1,2).<br>
4p=8 is the length of the latus rectum, so the two endpoints of the latus rectum are 2p=4 units above and below the focus, at (-1,6) and (-1,-2).<br>
A graph....<br>
{{{graph(400,400,-8,8,-8,8,sqrt(8(x+3))+2,-sqrt(8(x+3))+2)}}}<br>
vertex (-3,2)
focus (-1,2)
endpoints of latus rectum: (-1,-2) and (-1,6)
length of latus rectum: 8