The co-ordinates of the point S are (4, 0) and a point P has coordinates (x, y). Express `PS^(2)` in terms of x and y. Given that M is the foot of the perpendicular from P to the y-axis and that the point P moves so that lengths PS and PM are equal, prove that the locus of P is `8x=y^(2)+16`. Find the co-ordinates of one of the two points on the curve whose distance from S is 20 units.

Answer

To solve the problem step by step, we need to express the distance \( PS^2 \) in terms of \( x \) and \( y \), and then find the locus of point \( P \) such that \( PS = PM \). Finally, we will find the coordinates of the points on the curve whose distance from \( S \) is 20 units.

### Step 1: Express \( PS^2 \) in terms of \( x \) and \( y \)

Given the coordinates of point \( S \) are \( (4, 0) \) and point \( P \) has coordinates \( (x, y) \).

Using the distance formula, we can express \( PS^2 \) as follows:

\[ PS^2 = (x - 4)^2 + (y - 0)^2 \]

This simplifies to:

\[ PS^2 = (x - 4)^2 + y^2 \]

### Step 2: Find the coordinates of point \( M \)

Point \( M \) is the foot of the perpendicular from point \( P \) to the y-axis. Therefore, the coordinates of point \( M \) are \( (0, y) \).

### Step 3: Express \( PM^2 \)

Using the distance formula again, we can express \( PM^2 \):

\[ PM^2 = (x - 0)^2 + (y - y)^2 = x^2 \]

### Step 4: Set \( PS^2 \) equal to \( PM^2 \)

Since it is given that \( PS = PM \), we can set the squares equal to each other:

\[ PS^2 = PM^2 \]

Substituting the expressions we derived:

\[ (x - 4)^2 + y^2 = x^2 \]

### Step 5: Expand and simplify the equation

Expanding the left side:

\[ (x^2 - 8x + 16) + y^2 = x^2 \]

Now, subtract \( x^2 \) from both sides:

\[ -8x + 16 + y^2 = 0 \]

Rearranging gives:

\[ y^2 = 8x - 16 \]

### Step 6: Rearranging to find the locus equation

To express it in the standard form, we can rewrite it as:

\[ 8x = y^2 + 16 \]

This is the equation of the locus of point \( P \).

### Step 7: Find the coordinates of points on the curve whose distance from \( S \) is 20 units

The distance \( PS \) is given as 20 units. Therefore, we can set up the equation:

\[ PS^2 = 20^2 = 400 \]

Using our expression for \( PS^2 \):

\[ (x - 4)^2 + y^2 = 400 \]

### Step 8: Substitute \( y^2 \) from the locus equation

From the locus equation, we know:

\[ y^2 = 8x - 16 \]

Substituting this into the distance equation:

\[ (x - 4)^2 + (8x - 16) = 400 \]

### Step 9: Expand and simplify

Expanding the equation:

\[ (x^2 - 8x + 16) + 8x - 16 = 400 \]

This simplifies to:

\[ x^2 = 400 \]

### Step 10: Solve for \( x \)

Taking the square root gives:

\[ x = 20 \quad \text{or} \quad x = -20 \]

### Step 11: Find corresponding \( y \) values

Using the locus equation \( y^2 = 8x - 16 \):

1. For \( x = 20 \):

\[ y^2 = 8(20) - 16 = 160 - 16 = 144 \implies y = 12 \quad \text{or} \quad y = -12 \]

2. For \( x = -20 \):

\[ y^2 = 8(-20) - 16 = -160 - 16 = -176 \quad \text{(not possible)} \]

### Conclusion

The coordinates of one of the points on the curve whose distance from \( S \) is 20 units are:

\[ (20, 12) \quad \text{and} \quad (20, -12) \]