Question 1115791
{{{y^2-6y=-8x-25}}}
{{{y^2-6y+9=-8x-25+9}}}
{{{(y-3)^2=-8x-16}}}
{{{(y-3)^2=-8(x+2)}}}


vertex  (-2,3)


Distance between vertex and focus  p;
{{{-8=-4p}}}
{{{p=2}}}
Focus is two unit to the left of (-2,3)
Focus is at (-4,3).


LATUS RECTUM?
{{{y=3+- sqrt(-8(x+2))}}}
x coordinate of focus is  -4.
{{{y=3+- sqrt(-8(-4+2))}}}
{{{y=3+- 4}}}
{{{system(y=-1,or,7)}}}-------distance across the parabola at x=-4, is  8.
Latus Recturm is {{{highlight(8)}}}.