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

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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)   (Show Source): You can put this solution on YOUR website!


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



Use the distance formula to find 

Use the slope formula and the point-slope formula to find the
equation of AB which is 

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

d =  = distance from point (x1, y1) to line Ax+By+C=0









So we find the minimum value of  by using the
vertex formula.  The vertex formula for the x-coordinate of the
vertex of parabola  is .

So the x-coordinate of the parabola  is 

The minimum value is the y-coordinate of the vertex, which is


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



Edwin
Answer by ikleyn(52786)   (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)   (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:



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.


 

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