The normal at `(a,2a)` on `y^2 = 4ax` meets the curve again at `(at^2, 2at)`. Then the value of `t=`

Options

A. 1
B. 3
C. -1
D. -3

Answer

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

### Step 1: Identify the given point and the parabola
The problem states that we have a parabola given by the equation \( y^2 = 4ax \) and a point on the parabola at \( (a, 2a) \).

### Step 2: Determine the parameter \( t_1 \)
The point \( (a, 2a) \) can be expressed in terms of the parameter \( t_1 \) of the parabola. For the parabola \( y^2 = 4ax \), the coordinates of a point can be represented as \( (at_1^2, 2at_1) \).

From the given point \( (a, 2a) \):
- Comparing the y-coordinates:
\[ 2a = 2at_1 \implies t_1 = 1 \]

### Step 3: Use the relation between \( t_1 \) and \( t_2 \)
According to the properties of the parabola, the normal at a point \( (at_1^2, 2at_1) \) meets the parabola again at another point \( (at_2^2, 2at_2) \). The relation between \( t_1 \) and \( t_2 \) is given by:
\[ t_2 = -t_1 - \frac{2}{t_1} \]

Substituting \( t_1 = 1 \):
\[ t_2 = -1 - \frac{2}{1} = -1 - 2 = -3 \]

### Step 4: Conclusion
The value of \( t \) is equal to \( t_2 \), which we found to be:
\[ t = -3 \]

Thus, the final answer is:
\[ \boxed{-3} \]

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