Question 1181164


given:
 

Vertex at ({{{5}}},{{{-3}}}) =>{{{h=5}}} and {{{k=-3}}}
focus at ({{{6}}},{{{-3}}})
noticed that focus and vertex lie on the same horizontal line {{{y = -3}}}

parabola open to the right and the standard equation is:

{{{(y-k)^2=4p(x-h)}}} 

Distance from vertex to focus is {{{p =6-5= 1}}}

{{{(y-(-3))^2=4*1(x-5)}}} 
{{{(y+3)^2=4(x-5)}}} 



Parts of the Parabola
1. Vertex is at ({{{5}}},{{{-3}}})
2. Focus is at ({{{6}}},{{{-3}}})
3. Directrix: {{{x = 4}}}
4. Axis of Symmetry: {{{y=-3}}}
5. 
use ({{{6}}},{{{-3}}}),  E1, E1 will be where a line {{{x=6}}} intersect parabola

{{{(y+3)^2=4(6-5)  }}} ...plug in {{{x=6}}}

{{{(y+3)^2=4  }}}

{{{y+3= sqrt(4) }}}

{{{y+3= 2 }}} or {{{y+3= -2 }}}
{{{y= 2-3 }}} or {{{y= -2-3 }}}
{{{y= -1 }}} or {{{y= -5 }}}

E1=({{{6}}},{{{-1}}}), E1=({{{6}}},{{{-5}}})
6. Length of E1, E2 is distance between ({{{6}}},{{{-1}}}) and ({{{6}}},{{{-5}}}), and it is {{{4}}}


{{{drawing ( 600, 600, -10, 20, -10, 20,
circle(5,-3,.13),locate(5,-3,V(5,-3)),

circle(6,-3,.13),locate(6,-3,F(6,-3)),

circle(6,-1,.13),locate(6,-1,E1),
circle(6,-5,.13),locate(6,-5,E2),

blue(line(-5,-3,5,-3)), 
graph( 600, 600, -10, 20, -10, 20,sqrt(4(x-5))-3, - sqrt(4(x-5))-3, -3) ) }}}