Find the locus of the middle points of chords of a parabola which subtend a right angle at the vertex, and prove that these chords all pass through a fixed point on the axis of the curve.

Answer

To find the locus of the midpoints of chords of a parabola that subtend a right angle at the vertex, we will follow these steps:

### Step 1: Understand the Parabola
We start with the standard equation of the parabola:
\[ y^2 = 4ax \]
where \( a \) is a constant.

### Step 2: Parametric Representation
The points on the parabola can be represented parametrically as:
- Point A: \( (at_1^2, 2at_1) \)
- Point B: \( (at_2^2, 2at_2) \)

### Step 3: Midpoint of the Chord
The midpoint \( M \) of the chord AB can be calculated as:
\[ H = \frac{at_1^2 + at_2^2}{2} = \frac{a(t_1^2 + t_2^2)}{2} \]
\[ K = \frac{2at_1 + 2at_2}{2} = a(t_1 + t_2) \]

### Step 4: Condition for Right Angle
Since the chord subtends a right angle at the vertex, the slopes of OA and OB must satisfy the condition:
\[ m_1 \cdot m_2 = -1 \]
where:
\[ m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}, \quad m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2} \]
Thus,
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4 \]

### Step 5: Express \( t_1^2 + t_2^2 \)
Using the identity:
\[ (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1 t_2 \]
we can express \( t_1^2 + t_2^2 \) as:
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2 \]
Substituting \( t_1 t_2 = -4 \):
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 + 8 \]

### Step 6: Substitute into Midpoint Coordinates
Let \( s = t_1 + t_2 \):
\[ H = \frac{a((t_1 + t_2)^2 + 8)}{2} = \frac{a(s^2 + 8)}{2} \]
\[ K = as \]

### Step 7: Eliminate \( s \)
From \( K = as \), we have \( s = \frac{K}{a} \). Substituting this into the equation for \( H \):
\[ H = \frac{a\left(\left(\frac{K}{a}\right)^2 + 8\right)}{2} = \frac{K^2}{2a} + 4a \]

### Step 8: Rearranging to Find the Locus
Rearranging gives us:
\[ H - 4a = \frac{K^2}{2a} \]
Multiplying through by \( 2a \):
\[ 2a(H - 4a) = K^2 \]
Thus, we have:
\[ K^2 = 2aH - 8a^2 \]

### Step 9: Final Equation of the Locus
This can be rewritten as:
\[ K^2 = 2a(H - 4a) \]
This represents a parabola.

### Conclusion
The locus of the midpoints of the chords that subtend a right angle at the vertex of the parabola \( y^2 = 4ax \) is another parabola given by the equation:
\[ K^2 = 2a(H - 4a) \]
This shows that all such chords pass through the fixed point \( (4a, 0) \) on the axis of the parabola.