If `P(x_1,y_1),Q(x_2,y_2)` and `R(x_3, y_3)` are three points on `y^2 =4ax` and the normal at PQ and R meet at a point, then the value of `(x_1-x_2)/(y_3)+(x_2-x_3)/(y_1)+(x_3-x_1)/(y_2)=`

Options

A. 4a
B. 2a
C. a
D. 0

Explanation

To solve the problem step by step, we will analyze the given points on the parabola \( y^2 = 4ax \) and derive the required expression.

### Step 1: Understand the Equation of the Parabola
The equation of the parabola is given as \( y^2 = 4ax \). This means that for any point \( P(x_1, y_1) \), \( Q(x_2, y_2) \), and \( R(x_3, y_3) \) lying on the parabola, the coordinates must satisfy this equation.

### Step 2: Express \( x_1, x_2, x_3 \) in terms of \( y_1, y_2, y_3 \)
From the parabola's equation, we can express \( x_1, x_2, x_3 \) as follows:
- \( x_1 = \frac{y_1^2}{4a} \)
- \( x_2 = \frac{y_2^2}{4a} \)
- \( x_3 = \frac{y_3^2}{4a} \)

### Step 3: Use the Condition of Normal at Points
The problem states that the normals at points \( P \) and \( Q \) and at point \( R \) meet at a point. This implies that the sum of the ordinates of the feet of the normals drawn to the parabola is zero. Therefore, we have:
\[ y_1 + y_2 + y_3 = 0 \]
From this, we can express \( y_3 \) in terms of \( y_1 \) and \( y_2 \):
\[ y_3 = - (y_1 + y_2) \]

### Step 4: Substitute \( y_3 \) into the Expression
Now we need to evaluate the expression:
\[ \frac{x_1 - x_2}{y_3} + \frac{x_2 - x_3}{y_1} + \frac{x_3 - x_1}{y_2} \]
Substituting the values of \( x_1, x_2, x_3 \):
\[ = \frac{\frac{y_1^2}{4a} - \frac{y_2^2}{4a}}{- (y_1 + y_2)} + \frac{\frac{y_2^2}{4a} - \frac{y_3^2}{4a}}{y_1} + \frac{\frac{y_3^2}{4a} - \frac{y_1^2}{4a}}{y_2} \]
This simplifies to:
\[ = \frac{y_1^2 - y_2^2}{-4a(y_1 + y_2)} + \frac{y_2^2 - y_3^2}{4a y_1} + \frac{y_3^2 - y_1^2}{4a y_2} \]

### Step 5: Simplify Each Term
Using the identity \( a^2 - b^2 = (a - b)(a + b) \):
1. The first term becomes:
\[ = \frac{(y_1 - y_2)(y_1 + y_2)}{-4a(y_1 + y_2)} = \frac{-(y_1 - y_2)}{4a} \]
2. The second term using \( y_3 = - (y_1 + y_2) \):
\[ = \frac{y_2^2 - (- (y_1 + y_2))^2}{4a y_1} = \frac{y_2^2 - (y_1^2 + 2y_1y_2 + y_2^2)}{4a y_1} = \frac{-y_1^2 - 2y_1y_2}{4a y_1} \]
3. The third term becomes:
\[ = \frac{(- (y_1 + y_2))^2 - y_1^2}{4a y_2} = \frac{(y_1^2 + 2y_1y_2 + y_2^2) - y_1^2}{4a y_2} = \frac{2y_1y_2 + y_2^2}{4a y_2} \]

### Step 6: Combine All Terms
Combining all three terms, we will find that they sum to zero:
\[ \frac{-(y_1 - y_2)}{4a} + \frac{-y_1 - 2y_1y_2}{4a y_1} + \frac{2y_1y_2 + y_2^2}{4a y_2} = 0 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{0} \]