Question 1129284

Find the latus rectum, vertex and directrix of the parabola 

{{{y^2-8x-2y+17=0 }}}...first write it in vertex form

{{{(y^2-2y+b^2)-b^2-8x+17=0 }}}
{{{(y^2-2y+1^2)-1^2-8x+17=0 }}}
{{{(y-1)^2)-8(x-2)=0 }}}
{{{(y-1)^2)=8(x-2)=0 }}}=> compare to form {{{(y-k)^2=4p(x-h)}}}

=>{{{h=2}}} and {{{k=1}}} => vertex is at  ({{{2}}}, {{{1}}})
{{{4p=8}}}
{{{p=2}}}

focus: ({{{h+p}}},{{{k}}})= ({{{4}}},{{{ 1}}})

Equation of the directrix is   {{{x=-p}}}
{{{x-2=-2}}}
{{{x=2-2}}}
{{{x = 0}}}


The line segment that passes through the focus ({{{4}}},{{{ 1}}})  and is parallel to the directrix {{{x=0}}} is called the latus rectum. 
The {{{endpoints}}} of the latus rectum lie on the curve, and they are: 

({{{p}}},{{{2p}}}) and ({{{p}}},{{{-2p}}})

since {{{p=2}}}, we have

({{{2}}},{{{4}}}) and ({{{2}}},{{{-4}}})

distance between them is the length  of the latus rectum:

{{{d=sqrt((2-2)^2+(4-(-4))^2)}}}

{{{d=sqrt(0+(4+4)^2)}}}
{{{d=sqrt(8^2)}}}
{{{d=8}}}

so,  length of the latus rectum = {{{ 8}}}