Question 1131569
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If you use this form of the equation for a parabola<br>
{{{y = (1/(4p))(x-h)^2+k}}}<br>
Then the vertex is (h,k) and the focal width is 4p.<br>
So with the two given points on the parabola, and knowing the focal width is 8, we get two equations in h and k:<br>
(1) {{{0 = (1/8)(5-h)^2+k}}}
{{{0 = (1/8)(h^2-10h+25)+k}}}<br>
(2) {{{6 = (1/8)(9-h)^2+k}}}
{{{6 = (1/8)(h^2-18h+81)+k}}}<br>
Subtracting (1) from (2) eliminates k:<br>
{{{6 = (1/8)(-8h+56)}}}
{{{-8h+56 = 48}}}
{{{-8h = -8}}}
{{{h = 1}}}<br>
Substituting h=1 in (1) gives us k:<br>
{{{0 = (1/8)(5-1)^2+k}}}
{{{0 = 2+k}}}
{{{k = -2}}}<br>
ANSWER: An equation of the given parabola is {{{y = (1/8)(x-1)^2-2}}}<br>
A graph....  The vertex is (h,k) = (1,-2).  p=2 is the distance from the vertex to the focus, so the focus is (1,0); so you can see in the graph that the focal width is 8, with the parabola having x-intercepts -3 and 5.<br>
{{{graph(400,280,-10,10,-4,10,(1/8)(x-1)^2-2)}}}