The normal chord of a parabola `y^2= 4ax` at the point `P(x_1, x_1)` subtends a right angle at the

Options

A. focus
B. vertex
C. end of the latusrectum
D. none of these

Answer

To solve the problem, we need to analyze the properties of the normal chord of the parabola given by the equation \( y^2 = 4ax \) at the point \( P(x_1, y_1) \).

### Step-by-Step Solution:

1. **Identify the Point on the Parabola**:
The point \( P(x_1, y_1) \) lies on the parabola \( y^2 = 4ax \). Since \( P \) is given as \( (x_1, x_1) \), we can substitute \( y_1 = x_1 \) into the parabola's equation.
\[ (x_1)^2 = 4a(x_1) \]
This implies:
\[ x_1^2 = 4ax_1 \]

2. **Find the Normal at Point P**:
The slope of the tangent to the parabola at point \( P \) can be found using the derivative. The derivative of \( y^2 = 4ax \) gives us:
\[ \frac{dy}{dx} = \frac{2a}{y} \]
At point \( P(x_1, x_1) \):
\[ \frac{dy}{dx} = \frac{2a}{x_1} \]
The slope of the normal line is the negative reciprocal of the tangent slope:
\[ \text{slope of normal} = -\frac{x_1}{2a} \]

3. **Equation of the Normal Line**:
The equation of the normal line at point \( P(x_1, x_1) \) can be expressed using the point-slope form:
\[ y - x_1 = -\frac{x_1}{2a}(x - x_1) \]
Simplifying this, we get:
\[ y - x_1 = -\frac{x_1}{2a}x + \frac{x_1^2}{2a} \]
Rearranging gives:
\[ y = -\frac{x_1}{2a}x + \left(\frac{x_1^2}{2a} + x_1\right) \]

4. **Finding the Chord**:
The normal chord is the line segment of the normal that intersects the parabola at two points. For this specific case, we know that the normal chord subtends a right angle at the vertex of the parabola.

5. **Conclusion**:
It is a known property of parabolas that the normal chord at any point on the parabola subtends a right angle at the vertex. Therefore, the answer to the question is:
\[ \text{The normal chord subtends a right angle at the vertex of the parabola.} \]