Find the equation for the graph of the path of a point P(x,y) that moves such that P(x,y) is the same distance from the point (14,0) and the y-axis.
The focus is (14,0) and the directrix is x=0
SO, you have a parabola opening to the right, with 2p=14, and vertex at (7,0) halfway from directrix to focus:
y^2 = 28(x-7)
http://www.wolframalpha.com/input/?i=parabola+y%5E2+%3D+28(x-7)
To find the equation for the graph of the path of point P(x, y) that is equidistant from the point (14, 0) and the y-axis, we can follow these steps:
Step 1: Understand the problem
We need to find an equation that represents all the points P(x, y) that are equidistant from (14, 0) and the y-axis. In other words, the distance from P to (14, 0) should be equal to the distance from P to the y-axis.
Step 2: Find the distance formula
The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
Step 3: Determine the distance between P and (14, 0)
Using the distance formula, we can find the distance between P(x, y) and (14, 0) as follows:
d1 = √[(x - 14)^2 + (y - 0)^2]
= √[(x - 14)^2 + y^2]
Step 4: Determine the distance between P and the y-axis
To find the distance between P(x, y) and the y-axis, we need to consider that the y-axis is a vertical line passing through (0, 0). So the distance is simply the x-coordinate of point P:
d2 = |x|
Step 5: Set up the equation
Since we want both distances, d1 and d2, to be equal, we can set up the equation:
√[(x - 14)^2 + y^2] = |x|
Step 6: Square both sides of the equation
By squaring both sides of the equation, we eliminate the square root:
[(x - 14)^2 + y^2] = x^2
Step 7: Expand and simplify
Expanding and rearranging the above equation, we get:
x^2 - 28x + 196 + y^2 = x^2
Step 8: Cancel out x^2 terms
Since x^2 appears on both sides of the equation, we can cancel it out:
- 28x + 196 + y^2 = 0
This is the equation for the graph of the path of the point P(x, y) that is equidistant from (14, 0) and the y-axis.