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

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) (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):

which simplifies to:

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:

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:

Now we solve. Squaring both sides:

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:

Now, if we rewrite without an exponent:

and then use the property we get:

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, is just equal to:

Back to squaring the right side of:

which simplifies as follows:

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:

Reordering:

Multiplying both sides by -1 (because I prefer positive leading coefficients):

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:

Adding 12 to both sides of our equation:

Dividing both sides by 12:

The first fraction reduces:

which can be rewritten as:

From this we can see that the center is the point (0, 0), the and the . I'll leave the graph up to you.
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