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 Thanks

Question 824635: 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 very much:)
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
Let point P be (x, y).
Then the distance from P to (4, 0), using the distance formula, is:
or just
Since...

x = 1 is a vertical line; and
since distance from a point to a line is measured perpendicularly; and
since perpendicular to vertical is horizontal

.. the distance from point P to the line x = 1 is a horizontal distance. And horizontal distances are calculated by subtracting the x-coordinates.
So the displace from point P to the line x = 1 is:
(Note: Absolute value is used because distance should be positive. And without the absolute value x-1 would be negative if x < 1.)

We are now ready to 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 an equation. The translation might be easier to see if we reword the quoted sentence to: "The distance from point P to the point (4, 0) is twice as much as the distance from point P to the line x = 1."

Now we simplify. Squaring the left side should be simple. And squaring the right side is not hard, either, since x-1, after we square it, could not be negative. So the absolute value is no longer needed!

Now we put the equation in standard form (for conic sections). Subtracting the entire left side from both sides we get:

This is the equation for part a.

Looking at our equation, with no xy terms and with the x and y squared terms having opposite signs, we should recognize that it is the equation for a hyperbola.

To graph this hyperbola it will help to put it into the form for hyperbolas. Adding 12 to each side we get:

Dividing both sides by 16 (in order to turn it into a 1):

Reducing the the first fraction:

Rewriting the numerators to fit the form:

We can now "read" the key data:

With the x squared in front of the subtraction, this hyperbola has a horizontal transverse axis.
The "h" and "k" are zeros. So the center of the hyperbola is (0, 0) (aka the origin).
. This makes Since "a" is the distance from the center to the vertices on the transverse axis and since center is (0, 0), the vertices are (2, 0) and (-2, 0).
so b = 4.
With a = 2 and b = 4, the slopes of the asymptotes are: + = +2

I'll leave the graphing up to you. Here's what it should look like:

(Don't mind the different colors or the gaps (which are there only because algebra.com's graphing software doesn't handle nearly vertical parts of a graph well.)
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