PP' is a focal chord of the parabola `y^(2)=8x`. If the coordinates of P are (18,12), find the coordinates of P'

Answer

To find the coordinates of point P' given that PP' is a focal chord of the parabola \( y^2 = 8x \) and the coordinates of point P are (18, 12), we can follow these steps:

### Step 1: Identify the parameters of the parabola
The equation of the parabola is given as \( y^2 = 8x \). We can rewrite this in the standard form \( y^2 = 4ax \). Here, we can identify that \( 4a = 8 \), which gives us \( a = 2 \).

### Step 2: Use the coordinates of point P
The coordinates of point P are given as (18, 12). According to the properties of focal chords in a parabola, if P has coordinates \( (at^2, 2at) \), then we can equate:
- \( at^2 = 18 \)
- \( 2at = 12 \)

### Step 3: Solve for t
From the second equation \( 2at = 12 \):
\[ 2 \cdot 2 \cdot t = 12 \implies 4t = 12 \implies t = 3 \]

### Step 4: Substitute t back to find a
Now substitute \( t = 3 \) into the first equation:
\[ at^2 = 18 \implies 2 \cdot (3^2) = 18 \implies 2 \cdot 9 = 18 \]
This confirms that our value of \( t \) is correct.

### Step 5: Find the coordinates of P'
For a focal chord, the coordinates of P' can be found using the formulas:
- \( P' = (a/t'^2, -2a/t') \)
where \( t' \) is the negative reciprocal of \( t \) (since for focal chords, \( t \cdot t' = -1 \)). Thus:
\[ t' = -\frac{1}{t} = -\frac{1}{3} \]

### Step 6: Calculate the coordinates of P'
Now substitute \( t' = -\frac{1}{3} \) into the formulas for P':
\[ P' = \left( \frac{2}{(-\frac{1}{3})^2}, -\frac{2 \cdot 2}{-\frac{1}{3}} \right) \]
Calculating each component:
1. For the x-coordinate:
\[ \frac{2}{(-\frac{1}{3})^2} = \frac{2}{\frac{1}{9}} = 2 \cdot 9 = 18 \]
2. For the y-coordinate:
\[ -\frac{2 \cdot 2}{-\frac{1}{3}} = \frac{4}{\frac{1}{3}} = 4 \cdot 3 = 12 \]

### Final Coordinates of P'
Thus, the coordinates of point P' are:
\[ P' = (18, -12) \]

### Summary of the Solution
The coordinates of point P' are \( (18, -12) \).

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