If `(x_1,y_1),(x_2,y_2)" and "(x_3,y_3)` are the feet of the three normals drawn from a point to the parabola `y^2=4ax`, then `(x_1-x_2)/(y_3)+(x_2-x_3)/(y_1)+(x_3-x_1)/(y_2)` is equal to

Options

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

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{x_1 - x_2}{y_3} + \frac{x_2 - x_3}{y_1} + \frac{x_3 - x_1}{y_2} \]

where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the feet of the normals drawn from a point to the parabola \(y^2 = 4ax\).

### Step 1: Parametrize the Points on the Parabola
The points on the parabola can be parametrized using the parameter \(t\):
- For \(P(t_1)\): \(x_1 = at_1^2\), \(y_1 = 2at_1\)
- For \(Q(t_2)\): \(x_2 = at_2^2\), \(y_2 = 2at_2\)
- For \(R(t_3)\): \(x_3 = at_3^2\), \(y_3 = 2at_3\)

### Step 2: Write the Normals' Equations
The normal at a point \((x_i, y_i)\) on the parabola can be expressed as:
\[ y - y_i = -\frac{1}{2a} (x - x_i) \]
Thus, the normals at points \(P\), \(Q\), and \(R\) will meet at a point \((h, k)\).

### Step 3: Use the Condition for Normals
The normals from the point \((h, k)\) to the parabola satisfy the equation:
\[ at_1 + at_2 + at_3 = 2a - h \]
This gives us relationships among the parameters \(t_1, t_2, t_3\).

### Step 4: Calculate Each Term
Now we calculate each term in the expression:

1. **First Term**:
\[ \frac{x_1 - x_2}{y_3} = \frac{at_1^2 - at_2^2}{2at_3} = \frac{a(t_1^2 - t_2^2)}{2at_3} = \frac{t_1^2 - t_2^2}{2t_3} \]
Using the identity \(a^2 - b^2 = (a-b)(a+b)\):
\[ = \frac{(t_1 - t_2)(t_1 + t_2)}{2t_3} \]

2. **Second Term**:
\[ \frac{x_2 - x_3}{y_1} = \frac{at_2^2 - at_3^2}{2at_1} = \frac{t_2^2 - t_3^2}{2t_1} = \frac{(t_2 - t_3)(t_2 + t_3)}{2t_1} \]

3. **Third Term**:
\[ \frac{x_3 - x_1}{y_2} = \frac{at_3^2 - at_1^2}{2at_2} = \frac{t_3^2 - t_1^2}{2t_2} = \frac{(t_3 - t_1)(t_3 + t_1)}{2t_2} \]

### Step 5: Combine the Terms
Now we combine all three terms:
\[ \frac{(t_1 - t_2)(t_1 + t_2)}{2t_3} + \frac{(t_2 - t_3)(t_2 + t_3)}{2t_1} + \frac{(t_3 - t_1)(t_3 + t_1)}{2t_2} \]

### Step 6: Simplify the Expression
When we add these fractions, we can find a common denominator (which is \(2t_1t_2t_3\)) and combine the numerators. After simplification, we will find that all terms cancel out, leading to:

\[ = 0 \]

### Conclusion
Thus, the value of the expression is:

\[ \boxed{0} \]