Question 1190796


the equation of the parabola with latus rectum {{{ 6}}} joining ({{{ -4}}}, {{{1}}}) and ({{{2}}}, {{{1}}})

Length of latus rectum is {{{ 6}}}, so the leading coefficient is ±{{{1/6}}}

{{{y = (1/6)(x-h)^2+k}}}

calculate {{{h}}} and {{{k}}}

use ({{{ -4}}}, {{{1}}})

{{{1 = (1/6)(-4-h)^2+k}}}

{{{k = 1/6 (-h^2 - 8 h - 10)}}}...........eq.1


and ({{{2}}}, {{{1}}})

{{{1 = (1/6)(2-h)^2+k}}}

{{{k =  -h^2/6 + (2 h)/3 + 1/3}}}...........eq.2

then, from eq.1 and eq.2 we have

{{{1/6 (-h^2 - 8 h - 10)=-h^2/6 + (2 h)/3 + 1/3}}}

{{{-2h - 2 = 0}}}

{{{h=-1}}}

go to

{{{k = 1/6 (-h^2 - 8 h - 10)}}}...........eq.1, substitute {{{h}}}

{{{k = 1/6 (-(-1)^2 - 8 (-1) - 10)=-1/2}}}


and, your equation is:

{{{y = (1/6)(x+1)^2-1/2}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(-4,1,.12),locate(-4,1,p(-4,1)),
circle(2,1,.12),locate(2,1,p(2,1)),
green(line(-4,1,2,1)),
graph( 600, 600, -10, 10, -10, 10,(1/6)(x+1)^2-1/2)) }}}