Can someone PLEASE help me!!! Algebra 2 is KICKING my BUTT!!

I am so stuck on this question it aint even funny.

For each point, its distance from the point (4,0) is twice its distance from the line x= -4. (by the way this deals with conics...)


THANX TO anyone that helps!!

This will get you started.

The following is not 100% accurate, but I can not find my error:

Draw a picture with the xy axes. Draw the vertical line x=-4. Draw the point (4,0). Now pick any other point (x,y) and draw it.

The distance from (x,y) to x=-4 is the distance between (x,y) and (-4,y), which is x - (-4) = x + 4.

The distance between (x,y) and (4,0) is SQUAREROOT[ (x-4)^2 +(y-0)^2]. (This uses the formula that the distance between two points (x1,y1) and (x2,y2) is SQUAREROOT[ (x1-x2)^2 + (y1-y2)^2 ] ).

So we have 2 * (x + 4) = SQUAREROOT[ (x-4)^2 +(y-0)^2].

Square both sides: [ 2*(x+4) ]^2 = (x-4)^2 +(y)^2.

Then simplify the expression. But truthfully, I don't think I have done this 100% correctly.

Well, I haven't done Alg. 2 in a while, but I can give you the correct method for doing math problems in general. The following steps work for almost any problem, from algebra 1 to calculus
1) state what you know
2) state the relationships (in this case, the distance from a point to (4,0) is always twice the distance from that point to the line x=-4)
3) solve for what you need to get

The first two steps are VERY important. It helps organize all your thoughts into an accessible manner.

-A math nerd

given: some points exist (along a line) from (-4,y) {1/3 of the total distance}, to (4,0) {2/3 of the total distance}. [side A,side B,side C] where (measure of A)^2 + (measure of B)^2 = (measure of C)^2.

put in some points and see what type of graph you get.

(-4,0) exists on line B (-4,y)
(4,0) exists as point A
x(sub1)- x(sub2) = delta x
y(sub1) - y(sub2) = delta y
-4 - 4 = -8 => |-8| = 8 total distance in X.
0 = 0 = 0 => |0| = 0 total distance in y.
8 * 2/3 = 16 / 3 = 5 + 1/3.
4 - (5 + 1/3) = -4/3.
(-4/3 , 0) is the first point we plot.

using my theorem for finding all triangles (with all sides integer):
[8,6,10] i.e. 8 is distance in x, so distance in y must be six.
we plot points (-4,6), (-4,-6).
distance from (-4,6) to (4,0) is 10 by my integer triangle theorem.
distance from (-4,-6) to (4,0) is 10 by my integer triangle theorem.
10 * 2/3 = 6 +2/3 so the point we graph should be that far from (4,0).
8 * 2/3 = 5 + 1/3 => 4 - (5 + 1/3) = -4/3 = x.
6 * 2/3 = 4 - 0 = 4 = y
(-4/3 , 4)
check (5+1/3)^2 + 4^2 = 28 + 4/9 + 16 = 44+4/9 = 400/9 => 20/3 = 6+2/3.
so we did it right. Note: distance in a plane=(delta x^2 + delta y^2)^(1/2).
(-4/3. -4) also exists, so we graph it as well. (similar math).
all our points we have graphed exist on the line x = -4/3.

note:
my general triangle theorem for triangles with all sides integer,

starting with a difference of one between two sides of right triangle.
n ^2 - (n-1)^2 = n^2 - (n^2 - n - n + 1) = n^ 2 - n^2 + 2n - 1 = 2n -1 =
n + (n-1). Since a^2 + b^2 = c^2 => c^2 - b^2 = a^2 => a^2 = 2n -1 =>
a = (2n -1)^(1/2).
when a>= 3 .: a^2 = 2n +1 => 9 = 2n+1 => 10 =2n => n=5.
n = 5, n-1 = 4, third side of traingle = 3.
(k^2 + 1)/2 = c (where k >= 3 and odd. i.e. k = a ). c-1 = b. k = a.
(a = 2p+1 for all p => 1. ([(2n-1)^(1/2),n-1,n]. n=5. (a =2p +1 for all p => 0)
examples [3,4,5],[5,12,13],[7,24,25],[9,40,41], [k,((k^2-1)/2),(k^2+1)/2].
QED

difference of two between two sides of right triangle.
n^2 - (n-2)^2 => [(4n-4)^(1/2),n-2,n]. n=5. (a = 2p +4 for all p =>0).: k=a.
examples[4,3,5],[6,8,10], [k,((k^2)-4)/4,((k^2)+4)/4].

difference of three between two sides of right triangle.
n^2 - (n-3)^2 => [(6n-9)^(1/2),n-3,n] . n=15. (a = 6p + 9 for all p =>0).: k=a. examples[9,12,15],[15,36,39], [k,((k^2)-9)/6,(k^2+9)/6].

difference of four between two sides of right triangle.
n^2 - (n-4)^2 (from definition of polynomial expansion) implies
[(8n-16)^(1/2),n-4,n] n=10. (a = 4p + 8 for all p >= 0). K = a.
examples [8,6,10],[12,16,20], [k,((k^2)-16)/8,((k^2)+16)/8].

difference of five between two sides of right triangle.
n^2 - (n-5)^2 => [(10n-25)^(1/2),n-5,n] n=25. (a = 10p + 25 for all p>=0).:
k=a. examples [15,20,25],[25,60,65], [k,((k^2)-25)/10,((k^2)+25)/10].

difference of any integer value L between two sides of a right triangle.
[((n^2 - (n-L)^2)^(1/2),n-L,n] => [k,((k^2)-(L^2)/2L,((k^2)+(L^2)/2L].

notice as L increases the distance between the integers increases proving integer right triangles are infinite.

QED