SOLUTION: Given points A(0, 1), B(2, 5) and point P moving along the parabola y = x^2 + 4x + 7 …eq. 1, find the minimum value of area S of triangle PAB

Algebra ->  Triangles -> SOLUTION: Given points A(0, 1), B(2, 5) and point P moving along the parabola y = x^2 + 4x + 7 …eq. 1, find the minimum value of area S of triangle PAB      Log On


   



Question 1181254: Given points A(0, 1), B(2, 5) and point P moving along the parabola y = x^2 + 4x + 7 …eq. 1, find the minimum value of area S of triangle PAB
Found 3 solutions by Edwin McCravy, ikleyn, greenestamps:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!


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

matrix%281%2C3%2CArea%2Cof%2CPAB%29%22%22=%22%22S%22%22=%22%22%281%2F2%29%28base%29%28height%29%22%22=%22%22%281%2F2%29%28AB%29%28PQ%29

Use the distance formula to find AB%22%22=%22%222sqrt%285%29

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

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

d = abs%28Ax%5B1%5D%2BBy%5B1%5D%2BC%29%2Fsqrt%28A%5E2%2BB%5E2%29 = distance from point (x1, y1) to line Ax+By+C=0

PQ%22%22=%22%22abs%282x-y%2B1%29%2Fsqrt%282%5E2%2B%28-1%29%5E2%29%22%22=%22%22abs%282x-%28x%5E2%2B4x%2B7%29%2B1%29%2Fsqrt%284%2B1%29%22%22=%22%22abs%282x-x%5E2-4x-7%2B1%29%2Fsqrt%285%29

PQ%22%22=%22%22abs%28-2x-x%5E2-6%29%2Fsqrt%285%29%22%22=%22%22abs%282x%2Bx%5E2%2B6%29%2Fsqrt%285%29

matrix%281%2C3%2CArea%2Cof%2CPAB%29%22%22=%22%22%281%2F2%29%28AB%29%28PQ%29

matrix%281%2C3%2CArea%2Cof%2CPAB%29%22%22=%22%22S%22%22=%22%22%281%2F2%29%282sqrt%285%29%29%28abs%282x%2Bx%5E2%2B6%29%2Fsqrt%285%29%29%22%22=%22%22abs%28x%5E2%2B2x%2B6%29

So we find the minimum value of S+=+x%5E2%2B2x%2B6 by using the
vertex formula.  The vertex formula for the x-coordinate of the
vertex of parabola y=Ax%5E2%2BBx%2BC is -B%2F%282A%29.

So the x-coordinate of the parabola S+=+x%5E2%2B2x%2B6 is -%282%29%2F%282%281%29%29%22%22=%22%22-1

The minimum value is the y-coordinate of the vertex, which is
S+=+%28-1%29%5E2%2B2%28-1%29%2B6=1-2%2B6=5

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



Edwin

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.

This problem is advanced, which means that it is above the average high school level.

So, I will assume that my visitor has an adequate level in Math (otherwise, there is no sense in discussing).


            The idea is that the base of the triangle is just given:
            it is the segment AB, whose length you can easily calculate.

            So, our task is to find the point P on the parabola,
            which provides minimum height (i.e. altitude) of the triangle.


It is the same as to say that our task is to find the point P on the parabola, CLOSEST to the line through the points A and B.


I solved a TWIN problem TODAY under this link

https://www.algebra.com/algebra/homework/Length-and-distance/Length-and-distance.faq.question.1181253.html


Having this solution as a TEMPLATE, complete the solution to this problem to the end on your own.


-----------------


Oops, it looks like Edwin just made everything I described in my post.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Points A and B determine the line with equation y=2x+1.

Here is a graph of that line and the given quadratic:

graph%28400%2C400%2C-5%2C5%2C-1%2C10%2C2x%2B1%2Cx%5E2%2B4x%2B7%29

Using AB as the base of the triangle, with P somewhere on the parabola, the minimum area of triangle PAB will be when the height of the triangle is smallest; that will happen when P is at the point on the parabola which is closest to the line. And that will be where the tangent to the parabola has the same slope as the line.

y=x^2+4x+7
y'=2x+4

The slope of the line is 2; find the point on the parabola where the slope of the tangent is 2:

2x+4=2
2x=-2
x=-1
y=1-4+7=4

The area of PAB is minimum when P is (-1,4).

Then use the "shoelace" method to find the area of the triangle with vertices (0,1), (2,5), and (-1,4).

0 1
X
2 5
/ X \
2 -1 4 0
/ X \
-5 0 1 8
/ \
0 -1
--- ---
-3 7
Area = (1/2)(7-(-3)) = 5

ANSWER: The minimum area of PAB is 5.