SOLUTION: A point P moves such that it is always twice as far from the points (4,0) as it is on the line x = 1 a) find the equation of the locus b) identify the locus and graph Thank

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A point P moves such that it is always twice as far from the points (4,0) as it is on the line x = 1 a) find the equation of the locus b) identify the locus and graph Thank      Log On


   

Question 823774: A point P moves such that it is always twice as far from the points (4,0) as it is on the line x = 1
a) find the equation of the locus
b) identify the locus and graph
Thanks so much in advance:)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let (x, y) be the point P. Then the distance from point P to (4, 0) would be (using the distance formula):
sqrt%28%28x-4%29%5E2%2B%28y-0%29%5E2%29
which simplifies to:
sqrt%28x%5E2-8x%2B16%2By%5E2%29

The distance between a point and a line is always measured perpendicularly. Since the given line x = 1 is a vertical line, we will measure the distance from point P to the line x = 1 horizontally. A horizontal distance is simply the difference between the x-coordinates. The x-coordinates of all the points on the line x = 1 are 1's. Since we do not know if the x-coordinate of P is greater than or less than 1 we do not know which to put first in the difference. This problem can be solved using absolute value. This makes the distance from point P to the line x = 1:
abs%28x-1%29

With the two expressions for the distances we can translate "A point P moves such that it is always twice as far from the points (4,0) as it is on the line x = 1" into the equation:
sqrt%28x%5E2-8x%2B16%2By%5E2%29+=+2%2Aabs%28x-1%29

Now we solve. Squaring both sides:
x%5E2-8x%2B16%2By%5E2+=+%282%2Aabs%28x-1%29%29%5E2
x%5E2-8x%2B16%2By%5E2+=+%282%29%5E2%2A%28abs%28x-1%29%29%5E2
Squaring the 2 is simple. But how do you square an absolute value? The answer may be obvious to you. If not then a basic property of absolute values can help:
abs%28a%29%2Aabs%28b%29+=+abs%28a%2Ab%29
Now, if we rewrite %28abs%28x-1%29%29%5E2 without an exponent:
abs%28x-1%29%2Aabs%28x-1%29
and then use the property we get:
abs%28%28x-1%29%28x-1%29%29
or
abs%28%28x-1%29%5E2%29
Now we have the absolute value of a perfect square. But perfect squares can never be negative. And since the absolute value of a non-negative is simply itself, abs%28%28x-1%29%5E2%29 is just equal to:
%28x-1%29%5E2

Back to squaring the right side of:
x%5E2-8x%2B16%2By%5E2+=+%282%29%5E2%2A%28abs%28x-1%29%29%5E2
which simplifies as follows:
x%5E2-8x%2B16%2By%5E2+=+4%2A%28x-1%29%5E2
x%5E2-8x%2B16%2By%5E2+=+4%2A%28x%5E2-2x%2B1%29
x%5E2-8x%2B16%2By%5E2+=+4x%5E2-8x%2B4%29

Last of all we put it in general standard form. For this we gather all the terms on one side. Subtracting the entire right side from both sides we get:
-3x%5E2%2B12%2By%5E2+=+0
Reordering:
-3x%5E2%2By%5E2%2B12+=+0
Multiplying both sides by -1 (because I prefer positive leading coefficients):
3x%5E2-y%5E2-12+=+0

At this point we should recognize the equation as the equation of a hyperbola. To help you graph it, let's put it in the standard form for hyperbolas:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1
Adding 12 to both sides of our equation:
3x%5E2-y%5E2=12
Dividing both sides by 12:
3x%5E2%2F12-y%5E2%2F12=1
The first fraction reduces:
x%5E2%2F4-y%5E2%2F12=1
which can be rewritten as:
%28x-0%29%5E2%2F4-%28y-0%29%5E2%2F12=1

From this we can see that the center is the point (0, 0), the a%5E2+=+4 and the b%5E2+=+12. I'll leave the graph up to you.