If the normal at `t_(1) ` on the parabola `y^(2)=4ax` meet it again at `t_(2)` on the curve then `t_(1)(t_(1)+t_(2))+2` = ?

Options

A. 0
B. 1
C. `t_(1)`
D. `t_(2)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Parabola
The given parabola is \( y^2 = 4ax \). This is a standard form of a parabola that opens to the right.

**Hint:** Recall the standard form of a parabola and its properties.

### Step 2: Equation of the Normal at \( t_1 \)
The coordinates of a point on the parabola corresponding to the parameter \( t_1 \) are \( (at_1^2, 2at_1) \). The equation of the normal at this point is given by:
\[ y - 2at_1 = -\frac{1}{2at_1}(x - at_1^2) \]
This simplifies to:
\[ y + 2at_1 = -\frac{1}{2at_1}x + at_1^2 \]
Rearranging gives:
\[ x + 2at_1 \cdot 2at_1 = 0 \implies x + 2at_1 + at_1^3 = 0 \]

**Hint:** Remember how to derive the equation of the normal line from a point on the parabola.

### Step 3: Finding the Intersection with the Parabola Again
The normal meets the parabola again at \( t_2 \), which corresponds to the point \( (at_2^2, 2at_2) \). Substituting these coordinates into the normal equation gives:
\[ 2at_2 + at_2^2 t_1 = 2at_1 + at_1^3 \]

**Hint:** Substitute the coordinates of the second intersection point into the normal equation.

### Step 4: Rearranging the Equation
Rearranging the equation:
\[ 2at_2 - 2at_1 = at_1^3 - at_2^2 t_1 \]
Factoring out \( 2a \) and \( at_1 \) gives:
\[ 2(t_2 - t_1) = t_1(t_1^2 - t_2^2) \]

**Hint:** Factor out common terms from both sides of the equation.

### Step 5: Using the Difference of Squares
Using the difference of squares, we can rewrite \( t_1^2 - t_2^2 \) as \( (t_1 - t_2)(t_1 + t_2) \):
\[ 2(t_2 - t_1) = t_1(t_1 - t_2)(t_1 + t_2) \]
Dividing both sides by \( (t_2 - t_1) \) (assuming \( t_2 \neq t_1 \)):
\[ 2 = t_1(t_1 + t_2) \]

**Hint:** Remember the properties of the difference of squares.

### Step 6: Finding the Required Expression
We need to find \( t_1(t_1 + t_2) + 2 \):
\[ t_1(t_1 + t_2) + 2 = 2 + 2 = 0 \]

**Hint:** Substitute the value you found into the expression you need to evaluate.

### Final Answer
Thus, the required value is:
\[ \boxed{0} \]