Question 1044782
Find the Parabola properties and graph

given:
{{{y^2+6x+6y=39}}}

Vertex:
first write your equation in vertex form {{{x-h=a(y-k)^2}}} (since y squared) where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the vertex
{{{6x=-y^2-6y+39}}}
{{{x=-y^2/6-6y/6+39/6}}}
{{{x=-(1/6)(y^2+6y)+13/2}}}
{{{x=-(1/6)(y^2+6y+b^2)-(-1/6)b^2+13/2}}}
{{{x=-(1/6)(y^2+6y+3^2)-(-1/6)3^2+13/2}}}
{{{x=-(1/6)(y+3)^2-(-1/6)9+13/2}}}
{{{x=-(1/6)(y+3)^2+3/2+13/2}}}
{{{x=-(1/6)(y+3)^2+16/2}}}
{{{x=-(1/6)(y+3)^2+8}}}
{{{k=-3}}} and {{{h=8}}}
vertex is at ( {{{8}}}, {{{-3}}})

Since the value of a is negative, the parabola opens left.

Focus: 
The focus of a parabola can be found by adding p to the x-coordinate h if the parabola opens left or right.
({{{h+p}}},{{{k}}})
Find p, the distance from the vertex to the focus.

Find {{{p}}}, the distance from the vertex to a focus of the parabola by using the following formula.
{{{p=1/4a}}}
since {{{a=-1/6}}}, we have
{{{p=1/(4(-1/6))}}}
{{{p=-6/4}}}
{{{p=-3/2}}}
({{{8-3/2}}},{{{-3}}})

({{{16/2-3/2}}},{{{-3}}})

 ( {{{13/2}}}, {{{-3}}})->focus

Directrix:
The directrix of a parabola is the vertical line found by subtracting p from the x-coordinate h
of the vertex if the parabola opens left or right.
{{{x=h-p}}}
{{{x=8-(-3/2)}}}
{{{x=16/2+3/2}}}
{{{x = 19/2}}}

Axis of symmetry:
Find the axis of symmetry by finding the line that passes through the vertex and the focus.
vertex is at ( {{{8}}}, {{{-3}}})
 ( {{{13/2}}}, {{{-3}}})->focus
and that will be {{{y=-3}}}

Endpoints of latus retum:

length of latus rectum = {{{4p=4*(-3/2)=2(-3)=-6}}}

endpoints of the latus rectum:  ({{{h+p}}},{{{k-2p}}}) and ({{{h+p}}},{{{k+2p}}}) 
since {{{p=-3/2}}}, {{{k=-3}}} and {{{h=8}}} we have

 ({{{8+(-3/2)}}},{{{-3-2(-3/2)}}}) and ({{{8+(-3/2)}}},{{{-3+2(-3/2)}}}) 
 ({{{16/2-3/2}}},{{{-3+3}}}) and ({{{16/2-3/2}}},{{{-3-3}}}) 
 ({{{13/2}}},{{{0}}}) and ({{{13/2}}},{{{-6}}}) 


{{{drawing( 600, 600, -15, 15, -15, 15,
circle(8,-3,.13),locate(7.5,-2,V(8,-3)),circle(13/2,-3,.13),locate(13/2,-3,F(13/2,-3)),line(19/2,15,19/2,-15),
 graph( 600, 600, -15, 15, -15, 15, sqrt((x-8)/(-(1/6)))-3, -sqrt((x-8)/(-(1/6)))-3,-3)) }}}