S is the point (4,0) and M is the foot of the perpendicular drawn From a point to the y-axis.If P moves such that the distance PS and PM remain equal,find the locus of point P
the locus is a parabola, as you can see from the definitions of the parabola. The directrix is the line x=0.
Use what you know about the equation
y^2 = 4px
In this case, the focus is at (4,0), so the vertex has been shifted to (2,0) and the equation is
y^2 = 4p(x-2)
Now figure out the value of p and you're done.
I didn't get that...
You mean locus is shifted..... How that can be possible
To find the locus of point P, we need to determine the condition that ensures the distance between P and both points S and M are equal.
Let's start by finding the coordinates of point M, which is the foot of the perpendicular drawn from point P to the y-axis.
Since P is moving, let's assume its coordinates are (x, y). The foot of the perpendicular from P to the y-axis will have coordinates (0, y) since it lies on the y-axis.
Now, let's find the distance between points P and S. The distance formula between two points (x1, y1) and (x2, y2) is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Applying this formula, we have:
PS = sqrt((4 - x)^2 + (0 - y)^2) ... (1)
Next, let's find the distance between points P and M:
PM = sqrt((0 - x)^2 + (y - y)^2) ... (2)
Since the distance between PS and PM should be equal, we can set up an equation equating the two distances:
sqrt((4 - x)^2 + (0 - y)^2) = sqrt((0 - x)^2 + (y - y)^2)
Squaring both sides of the equation, we get:
(4 - x)^2 + y^2 = x^2 + y^2
Expanding and simplifying the terms, we have:
16 - 8x + x^2 + y^2 = x^2 + y^2
Simplifying further, we get:
16 - 8x = 0
Solving this equation for x, we find:
x = 2
Substituting this value of x back into equation (2), we find:
PM = sqrt((0 - 2)^2 + (y - y)^2)
PM = sqrt(4) = 2
So, for any point P on the locus, the distance between P and S (PS) will be sqrt(20), and the distance between P and M (PM) will be 2.
Therefore, the locus of point P is a circle centered at the point M (0, 0) with a radius of 2.