Question 1202880
Opens to the right, length of the latus rectum = 6, and passing through

(2/3, 0) and (- 1/2, 7)
<pre>
Parabola opening right has p > 0, equation

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

latus rectum = 4p = 6

Substitute (x,y) = {{{(matrix(1,3,2/3,",",0))}}}

{{{(0-k)^2=6(2/3-h)^""}}}
{{{k^2=4-6h)}}}
{{{6h=4-k^2}}}

Substitute (x,y) = {{{(matrix(1,3,-1/2,",",7))}}}

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

{{{49-14k+k^2=-3-6h
{{{6h=-52+14k-k^2)}}}

Equating expressions for 6h

{{{4-k^2=-52+14k-k^2}}}
{{{4=-52+14k}}}
{{{56=14k}}}
{{{4=k}}}
Substitute in {{{6h=4-k^2}}}
{{{6h=4-4^2}}}
{{{6h=4-16}}}
{{{6h=-12}}}
{{{h=-2}}}

Equation is:

{{{(y-4)^2=6(x-(-2)^"")}}}
{{{(y-4)^2=6(x+2^"")}}}

{{{drawing(5600/19,800,-3,4,-3,13,
graph(5600/19,800,-3,4,-3,13,4+sqrt(6x+12)),
graph(5600/19,800,-3,4,-3,13,4-sqrt(6x+12)),
circle(2/3,0,.1),locate(1/2,1,(matrix(1,3,2/3,",",0))),
circle(-1/2,7,.1),locate(-2.4,7.2,(matrix(1,3,-1/2,",",7))),
red(arc(0,4,4,-4,170,190))
)}}}

Edwin</pre>