If the normals at `(x_(1),y_(1))` and `(x_(2) ,y_(2))` on `y^(2) = 4ax` meet again on parabola then `x_(1)x_(2)+y_(1)y_(2)=`

Options

A. `4a^(2)`
B. `8a^(2)`
C. `12a^(2)`
D. `10a^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the points on the parabola
The given parabola is \( y^2 = 4ax \). We can represent the points \( (x_1, y_1) \) and \( (x_2, y_2) \) on the parabola using the parameterization:
- \( (x_1, y_1) = (t_1^2, 2t_1) \)
- \( (x_2, y_2) = (t_2^2, 2t_2) \)

### Step 2: Write the normals at the points
The equation of the normal to the parabola at a point \( (t^2, 2t) \) is given by:
\[ y - 2t = -\frac{1}{t}(x - t^2) \]
Thus, the normals at the points \( (x_1, y_1) \) and \( (x_2, y_2) \) can be expressed as:
- Normal at \( (x_1, y_1) \):
\[ y - 2t_1 = -\frac{1}{t_1}(x - t_1^2) \]

- Normal at \( (x_2, y_2) \):
\[ y - 2t_2 = -\frac{1}{t_2}(x - t_2^2) \]

### Step 3: Find the intersection point of the normals
Let the normals intersect at a point \( (x, y) \) on the parabola, which can be represented as \( (t^2, 2t) \).

From the normal at \( (x_2, y_2) \):
\[ y - 2t_2 = -\frac{1}{t_2}(x - t_2^2) \]
Substituting \( y = 2t \) and \( x = t^2 \):
\[ 2t - 2t_2 = -\frac{1}{t_2}(t^2 - t_2^2) \]
This simplifies to:
\[ 2(t - t_2) = -\frac{1}{t_2}(t^2 - t_2^2) \]

From the normal at \( (x_1, y_1) \):
\[ y - 2t_1 = -\frac{1}{t_1}(x - t_1^2) \]
Substituting \( y = 2t \) and \( x = t^2 \):
\[ 2t - 2t_1 = -\frac{1}{t_1}(t^2 - t_1^2) \]
This simplifies to:
\[ 2(t - t_1) = -\frac{1}{t_1}(t^2 - t_1^2) \]

### Step 4: Equate the two expressions
Setting the two expressions for \( t \) equal gives:
\[ -\frac{t^2 - t_2^2}{t_2} = -\frac{t^2 - t_1^2}{t_1} \]
Cross-multiplying leads to:
\[ t_1(t^2 - t_2^2) = t_2(t^2 - t_1^2) \]

### Step 5: Rearranging and simplifying
Rearranging gives:
\[ t_1 t^2 - t_1 t_2^2 = t_2 t^2 - t_2 t_1^2 \]
This can be rearranged to:
\[ t_1 t_2^2 - t_2 t_1^2 + t_2 t^2 - t_1 t^2 = 0 \]
Factoring gives:
\[ (t_1 t_2 - 2)(t_2 - t_1) = 0 \]
Thus, we have:
\[ t_1 t_2 = 2 \]

### Step 6: Calculate \( x_1 x_2 + y_1 y_2 \)
Now substituting \( t_1 t_2 = 2 \):
- \( x_1 = t_1^2 \)
- \( x_2 = t_2^2 \)
- \( y_1 = 2t_1 \)
- \( y_2 = 2t_2 \)

Thus:
\[ x_1 x_2 + y_1 y_2 = t_1^2 t_2^2 + (2t_1)(2t_2) = (t_1 t_2)^2 + 4(t_1 t_2) \]
Substituting \( t_1 t_2 = 2 \):
\[ = 2^2 + 4 \cdot 2 = 4 + 8 = 12 \]

### Final Answer
Thus, the value of \( x_1 x_2 + y_1 y_2 \) is \( \boxed{12} \).

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