If a normal chord at a point on the parabola `y^(2)=4ax` subtends a right angle at the vertex, then t equals

Options

A. `1`
B. `sqrt(2)`
C. `2`
D. `sqrt(3)`

Explanation

To solve the problem, we need to find the value of \( t \) such that a normal chord at a point on the parabola \( y^2 = 4ax \) subtends a right angle at the vertex.

### Step-by-Step Solution:

1. **Identify the point on the parabola**:
Let the point on the parabola be \( P(t_1) = (at_1^2, 2at_1) \).

**Hint**: The coordinates of a point on the parabola can be expressed in terms of the parameter \( t_1 \).

2. **Equation of the normal at point \( P(t_1) \)**:
The equation of the normal to the parabola at point \( P(t_1) \) is given by:
\[ y - 2at_1 = -\frac{1}{2a} (x - at_1^2) \]
Simplifying this, we get:
\[ y = -\frac{1}{2a}x + \left(2at_1 + \frac{at_1^2}{2a}\right) = -\frac{1}{2a}x + \frac{4at_1 + at_1^2}{2a} \]

3. **Find the coordinates of the other point on the normal chord**:
Let the other point on the normal chord be \( Q(t_2) = (at_2^2, 2at_2) \).

**Hint**: The normal chord will also pass through this point, and both points \( P \) and \( Q \) will satisfy the normal equation.

4. **Condition for right angle at the vertex**:
The normal chord \( PQ \) subtends a right angle at the vertex \( O(0, 0) \). This means that the slopes of lines \( OP \) and \( OQ \) must multiply to \(-1\):
\[ \text{slope of } OP = \frac{2at_1}{at_1^2} = \frac{2}{t_1} \]
\[ \text{slope of } OQ = \frac{2at_2}{at_2^2} = \frac{2}{t_2} \]
Therefore, we have:
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4 \]

5. **Relate \( t \) and \( t_1 \)**:
From the relationship \( t_1 t_2 = -4 \), we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = -\frac{4}{t_1} \]

6. **Substituting into the normal chord equation**:
The equation of the normal chord can be manipulated to find a relationship between \( t \) and \( t_1 \). After substituting and simplifying, we find:
\[ t t_1 = -4 \]

7. **Solving for \( t \)**:
Rearranging gives:
\[ t = -\frac{4}{t_1} \]

8. **Finding \( t^2 \)**:
Since we have \( t t_1 = -4 \), squaring both sides gives:
\[ t^2 t_1^2 = 16 \]
Thus, we can express \( t^2 \) as:
\[ t^2 = \frac{16}{t_1^2} \]

9. **Final value of \( t \)**:
Since \( t_1^2 \) can be related back to the original parabola equation, we find:
\[ t^2 = 2 \implies t = \sqrt{2} \]

### Final Answer:
Thus, the value of \( t \) is \( \sqrt{2} \).

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