Find the locus of the middle points of chords of the parabola which
are normal to the curve

Answer

To find the locus of the midpoints of chords of the parabola \( y^2 = 4ax \) that are normal to the curve, we can follow these steps:

### Step 1: Write the equation of the normal to the parabola
For a point \( (x_1, y_1) \) on the parabola \( y^2 = 4ax \), the equation of the normal is given by:
\[ xy_1 + 2ay = 2a + x_1y_1 \]

### Step 2: Define the midpoint of the chord
Let the midpoint of the chord be \( ( \alpha, \beta ) \). The equation of the chord with this midpoint can be expressed as:
\[ y - \beta = m(x - \alpha) \]
where \( m \) is the slope of the chord.

### Step 3: Relate the midpoint to the normal
The normal at the point \( (x_1, y_1) \) can also be expressed in terms of the midpoint \( ( \alpha, \beta ) \). The slope of the normal is given by:
\[ m_{\text{normal}} = -\frac{y_1}{2a} \]
Thus, the equation of the normal can be rewritten as:
\[ y - \beta = -\frac{y_1}{2a}(x - \alpha) \]

### Step 4: Use the condition for the midpoint
The midpoint \( ( \alpha, \beta ) \) must satisfy both the normal equation and the chord equation. Therefore, we can set up the equations:
1. From the normal:
\[ \alpha y_1 + 2a\beta = 2a + x_1y_1 \]
2. From the chord:
\[ y - \beta = m(x - \alpha) \]

### Step 5: Substitute and simplify
Substituting \( y_1 \) and \( x_1 \) in terms of \( \alpha \) and \( \beta \) into the parabola equation \( y^2 = 4ax \):
\[ \beta^2 = 4a\alpha \]

### Step 6: Find the relationship between \( \alpha \) and \( \beta \)
From the normal equation, we can express \( x_1 \) and \( y_1 \) in terms of \( \alpha \) and \( \beta \):
\[ y_1 = -\frac{4a^2}{\beta} \]
Substituting this into the normal equation gives us:
\[ \beta^2 = 4a\alpha + 8a^2 \]

### Step 7: Rearranging the equation
Rearranging gives us the final equation:
\[ \beta^2 - 2a\alpha + 4a^2 + 8a^4 = 0 \]

### Conclusion
Thus, the locus of the midpoints of the chords of the parabola that are normal to the curve is given by:
\[ \beta^2 - 2a\alpha + 4a^2 + 8a^4 = 0 \]