The normal at a point `P (36,36)` on the parabola `y^(2)=36x` meets the parabola again at a point Q. Find the coordinates of Q.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given point on the parabola
The point \( P(36, 36) \) lies on the parabola defined by the equation \( y^2 = 36x \). We can express the coordinates of any point on the parabola in terms of a parameter \( t \).

### Step 2: Find the parameter \( t_1 \)
For the parabola \( y^2 = 36x \), we can rewrite it in the standard form \( y^2 = 4ax \) where \( a = 9 \). The general point on the parabola can be expressed as:
\[ (9t^2, 18t) \]
We need to find \( t_1 \) such that:
\[ 9t_1^2 = 36 \quad \text{and} \quad 18t_1 = 36 \]
From \( 18t_1 = 36 \):
\[ t_1 = 2 \]

### Step 3: Find the coordinates of point \( P \)
Substituting \( t_1 = 2 \) into the general point formula:
\[ P(9(2^2), 18(2)) = P(36, 36) \]
This confirms that \( P(36, 36) \) is indeed on the parabola.

### Step 4: Find the parameter \( t_2 \) for the normal
The parameter \( t_2 \) for the normal at point \( P \) is given by:
\[ t_2 = -t_1 - 2 = -2 - 2 = -4 \]

### Step 5: Find the coordinates of point \( Q \)
Now we can find the coordinates of point \( Q \) using the parameter \( t_2 \):
\[ Q(9t_2^2, 18t_2) = Q(9(-4)^2, 18(-4)) \]
Calculating these:
\[ Q(9 \times 16, -72) = Q(144, -72) \]

### Final Answer
The coordinates of point \( Q \) are:
\[ \boxed{(144, -72)} \]

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