Find the co-ordinates of the points lying on parabola `x^(2)=12y` whose focal distance is 15.

Answer

To solve the problem of finding the coordinates of the points lying on the parabola \( x^2 = 12y \) whose focal distance is 15, we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given equation of the parabola is \( x^2 = 12y \). We can compare this with the standard form of a parabola \( x^2 = 4ay \).

From the equation \( x^2 = 12y \), we can see that \( 4a = 12 \).

### Step 2: Calculate the value of \( a \)
To find \( a \), we divide both sides of the equation by 4:
\[ a = \frac{12}{4} = 3 \]

### Step 3: Determine the coordinates of the focus and the directrix
For the parabola \( x^2 = 4ay \):
- The focus is located at \( (0, a) \).
- The directrix is given by the equation \( y = -a \).

Substituting \( a = 3 \):
- The coordinates of the focus \( S \) are \( (0, 3) \).
- The equation of the directrix is \( y = -3 \).

### Step 4: Use the definition of focal distance
The focal distance \( d \) from any point \( P(x, y) \) on the parabola to the focus \( S(0, 3) \) is given by:
\[ d = \sqrt{(x - 0)^2 + (y - 3)^2} \]
We are given that this focal distance is 15, so we set up the equation:
\[ \sqrt{x^2 + (y - 3)^2} = 15 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides, we get:
\[ x^2 + (y - 3)^2 = 15^2 \]
\[ x^2 + (y - 3)^2 = 225 \]

### Step 6: Expand the equation
Expanding \( (y - 3)^2 \):
\[ x^2 + (y^2 - 6y + 9) = 225 \]
\[ x^2 + y^2 - 6y + 9 = 225 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ x^2 + y^2 - 6y + 9 - 225 = 0 \]
\[ x^2 + y^2 - 6y - 216 = 0 \]

### Step 8: Complete the square for the \( y \) terms
To complete the square for \( y \):
\[ y^2 - 6y = (y - 3)^2 - 9 \]
Substituting this back into the equation:
\[ x^2 + (y - 3)^2 - 9 - 216 = 0 \]
\[ x^2 + (y - 3)^2 = 225 \]

### Step 9: Identify the points on the circle
This equation represents a circle centered at \( (0, 3) \) with a radius of 15. The points \( (x, y) \) that satisfy this equation will be the points on the parabola that are 15 units away from the focus.

### Step 10: Solve for \( y \)
From the equation \( (y - 3)^2 = 225 - x^2 \):
\[ y - 3 = \pm \sqrt{225 - x^2} \]
Thus:
\[ y = 3 \pm \sqrt{225 - x^2} \]

### Conclusion
The coordinates of the points lying on the parabola \( x^2 = 12y \) whose focal distance is 15 are given by:
\[ (x, y) = \left(x, 3 + \sqrt{225 - x^2}\right) \quad \text{and} \quad \left(x, 3 - \sqrt{225 - x^2}\right) \]
for \( -15 \leq x \leq 15 \).