The normal drawn at a point `(at_(1)^2 2at_1)` of the parabola `y^2 = 4ax` meets it again in the point `(at_2^2,2at_2)`, then

Options

A. `t_1=2t_2`
B. `t_1^2=2t_2`
C. `t_1.t_2=1`
D. `t_1^2+t_1t_2+2=0`

Answer

To solve the problem, we need to find the relationship between \( t_1 \) and \( t_2 \) when the normal at the point \( (at_1^2, 2at_1) \) on the parabola \( y^2 = 4ax \) meets the parabola again at the point \( (at_2^2, 2at_2) \).

### Step-by-Step Solution:

1. **Identify the point on the parabola**:
The point on the parabola is given as \( P(at_1^2, 2at_1) \).

2. **Write the equation of the normal**:
The equation of the normal at point \( P \) is given by:
\[ y + t_1 x = 2at_1 + at_1^3 \]
Rearranging this, we get:
\[ y - 2at_1 = -t_1 x + at_1^3 \]

3. **Substitute the point \( Q(at_2^2, 2at_2) \) into the normal equation**:
The normal must pass through the point \( Q(at_2^2, 2at_2) \). Substituting \( x = at_2^2 \) and \( y = 2at_2 \) into the normal equation, we have:
\[ 2at_2 - 2at_1 = -t_1(at_2^2) + at_1^3 \]

4. **Simplify the equation**:
Rearranging gives:
\[ 2a(t_2 - t_1) = -at_1 t_2^2 + at_1^3 \]
Dividing through by \( a \) (assuming \( a \neq 0 \)):
\[ 2(t_2 - t_1) = -t_1 t_2^2 + t_1^3 \]

5. **Rearranging the equation**:
Rearranging gives:
\[ t_1 t_2^2 + 2(t_2 - t_1) + t_1^3 = 0 \]
This can be rewritten as:
\[ t_1^2 + t_1 t_2 + 2 = 0 \]

6. **Conclusion**:
The relationship between \( t_1 \) and \( t_2 \) is given by:
\[ t_1^2 + t_1 t_2 + 2 = 0 \]