Question 1204031



If the parabola is vertical, the equation of the parabola is of the form 

.{{{ y = a(x - h)^2 + k}}}

Focus ({{{1}}},{{{-1}}})= ({{{h}}}, {{{k + 1/(4a) }}})=> {{{h=1}}}, {{{k + 1/(4a)=-1}}}=>{{{k=-1-1/(4a)}}}

Directrix have an equation:

 {{{y = k - 1/(4a)}}}=>{{{y=-1-1/(4a)- 1/(4a)}}}=>{{{y=-1/(2 a) - 1}}}

passes through the point ({{{-31}}},{{{59}}})


{{{59 = a(-31 - 1)^2 + -1-1/(4a)}}}

{{{a = -1/256}}}

or

{{{a=1/16}}}


since given that parabola opens up, we need {{{a=1/16}}}

=>{{{k=-1-1/(4(1/16))=-5}}}

{{{y=-1/(2(1/16)) - 1=-9}}}


a. The directrix will have an equation of

{{{y=-9}}}

b. The equation of the parabola will be?

{{{y = (1/16)(x - 1)^2 -5}}}



{{{ drawing( 600, 600, -60, 60, -60, 60,
circle(1,-1,.25),locate (1,-1,F(1,-1)),
circle(-31,59,.25),locate (-31,59,p(-31,59)),

graph( 600, 600, -60, 60, -60, 60, (1/16)(x - 1)^2 -5,-9)) }}}