Question 1018044
{{{7y=-5x^2+5x+4}}}
{{{7y=-5(x^2-x)+4}}}
{{{7y=-5(x^2-x+1/4-1/4)+4}}}, completing the square using {{{1/4}}};
which, through a few more steps, will give
as the standard form, {{{highlight(y=-(5/7)(x-1/2)^2+11/28)}}}
Telling you that the vertex, being a maximum, is at  VERTEX:  (1/2, 11/28).


The equation in an equivalent form is also  {{{highlight((7/5)y=-(x-1/2)^2+11/20)}}}.
Comparing to {{{4py=-(x-1/2)^2+11/20}}}, the value of p is the distance between the FOCUS and the vertex.
-
{{{4p=7/5}}}
{{{highlight_green(p=7/20)}}}.


The focus, being below the vertex for this example,  will have x=1/2 but {{{y=11/28-7/20=(5/5)(11/28)-(7/7)(7/20)}}}
{{{y=(55-49)/(140)}}}
{{{y=6/140}}}
{{{y=3/70}}}
FOCUS:  ( 1/2, 3/70 ).