Question 1181254
<pre>
{{{drawing(3200/9,400,-5,3,-2,7,
locate(.7,2.4,Q),
locate(-4.2,4.2,"P(x,y)"),circle(-3,4,.08),
locate(.1,1.2,"A(0,1)"), locate(2.1,5.2,"B(2,5)"),
graph(3200/9,400,-5,3,-2,7,x^2+4x+7), line(0,1,2,5),
blue(line(-3,4,0,1),line(-3,4,2,5)), green(line(.6,2.2,-3,4)) )}}}

Let P(x,y) be an arbitrary point along the parabola.

{{{matrix(1,3,Area,of,PAB)}}}{{{""=""}}}{{{S}}}{{{""=""}}}{{{(1/2)(base)(height)}}}{{{""=""}}}{{{(1/2)(AB)(PQ)}}}

Use the distance formula to find {{{AB}}}{{{""=""}}}{{{2sqrt(5)}}}

Use the slope formula and the point-slope formula to find the
equation of AB which is {{{2x - y + 1 = 0}}}

Use the perpendicular distance from point to line formula to find PQ, 
which is:

d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}} = distance from point (x<sub>1</sub>, y<sub>1</sub>) to line Ax+By+C=0

{{{PQ}}}{{{""=""}}}{{{abs(2x-y+1)/sqrt(2^2+(-1)^2)}}}{{{""=""}}}{{{abs(2x-(x^2+4x+7)+1)/sqrt(4+1)}}}{{{""=""}}}{{{abs(2x-x^2-4x-7+1)/sqrt(5)}}}

{{{PQ}}}{{{""=""}}}{{{abs(-2x-x^2-6)/sqrt(5)}}}{{{""=""}}}{{{abs(2x+x^2+6)/sqrt(5)}}}

{{{matrix(1,3,Area,of,PAB)}}}{{{""=""}}}{{{(1/2)(AB)(PQ)}}}

{{{matrix(1,3,Area,of,PAB)}}}{{{""=""}}}{{{S}}}{{{""=""}}}{{{(1/2)(2sqrt(5))(abs(2x+x^2+6)/sqrt(5))}}}{{{""=""}}}{{{abs(x^2+2x+6)}}}

So we find the minimum value of {{{S = x^2+2x+6}}} by using the
vertex formula.  The vertex formula for the x-coordinate of the
vertex of parabola {{{y=Ax^2+Bx+C}}} is {{{-B/(2A)}}}.

So the x-coordinate of the parabola {{{S = x^2+2x+6}}} is {{{-(2)/(2(1))}}}{{{""=""}}}{{{-1}}}

The minimum value is the y-coordinate of the vertex, which is
{{{S = (-1)^2+2(-1)+6=1-2+6=5}}}

So the minimum area S of triangle PAB is S = 5
 
The original drawing should look like this:

{{{drawing(3200/9,400,-5,3,-2,7,
locate(1.1,3.2,"Q(1,3)"),
locate(-2.3,4.2,"P(-1,4)"),circle(-1,4,.08),
locate(.1,1.2,"A(0,1)"), locate(2.1,5.2,"B(2,5)"),
graph(3200/9,400,-5,3,-2,7,x^2+4x+7), line(0,1,2,5),
blue(line(-1,4,0,1),line(-1,4,2,5)), green(line(1,3,-1,4)) )}}}

Edwin</pre>