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 evaluate 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: Parametric Representation of Points
We can represent the points on the parabola using the parametric equations:
- For point P: \((x_1, y_1) = (at_1^2, 2at_1)\)
- For point Q: \((x_2, y_2) = (at_2^2, 2at_2)\)
- For point R: \((x_3, y_3) = (at_3^2, 2at_3)\)

### Step 2: Substitute the Parametric Values
Substituting these values into the expression:

\[ \frac{at_1^2 - at_2^2}{2at_3} + \frac{at_2^2 - at_3^2}{2at_1} + \frac{at_3^2 - at_1^2}{2at_2} \]

### Step 3: Simplify Each Term
We can factor out \(a\) from the numerator and \(2a\) from the denominator:

\[ = \frac{a(t_1^2 - t_2^2)}{2at_3} + \frac{a(t_2^2 - t_3^2)}{2at_1} + \frac{a(t_3^2 - t_1^2)}{2at_2} \]

This simplifies to:

\[ = \frac{t_1^2 - t_2^2}{2t_3} + \frac{t_2^2 - t_3^2}{2t_1} + \frac{t_3^2 - t_1^2}{2t_2} \]

### Step 4: Use the Difference of Squares
Using the identity \(a^2 - b^2 = (a - b)(a + b)\), we can rewrite each term:

\[ = \frac{(t_1 - t_2)(t_1 + t_2)}{2t_3} + \frac{(t_2 - t_3)(t_2 + t_3)}{2t_1} + \frac{(t_3 - t_1)(t_3 + t_1)}{2t_2} \]

### Step 5: Apply the Condition for Conormal Points
From the properties of conormal points, we know that:

\[ t_1 + t_2 + t_3 = 0 \]

This implies \(t_1 + t_2 = -t_3\), \(t_2 + t_3 = -t_1\), and \(t_3 + t_1 = -t_2\).

### Step 6: Substitute Back into the Expression
Substituting these back into our expression gives:

\[ = \frac{(t_1 - t_2)(-t_3)}{2t_3} + \frac{(t_2 - t_3)(-t_1)}{2t_1} + \frac{(t_3 - t_1)(-t_2)}{2t_2} \]

### Step 7: Simplify Each Term
This simplifies to:

\[ = \frac{-(t_1 - t_2)}{2} + \frac{-(t_2 - t_3)}{2} + \frac{-(t_3 - t_1)}{2} \]

### Step 8: Combine the Terms
Combining these terms leads to:

\[ = \frac{-(t_1 - t_2 + t_2 - t_3 + t_3 - t_1)}{2} = \frac{-0}{2} = 0 \]

### Final Answer
Thus, the value of the expression is:

\[ \frac{x_1 - x_2}{y_3} + \frac{x_2 - x_3}{y_1} + \frac{x_3 - x_1}{y_2} = 0 \]