Find the locus of the middle points of chords of the parabola which
pass through the fixed point (h,k)

Answer

To find the locus of the midpoints of the chords of the parabola \( y^2 = 4ax \) that pass through the fixed point \( (h, k) \), we can follow these steps:

### Step 1: Define the midpoint
Let the midpoint of the chord be \( ( \alpha, \beta ) \).

### Step 2: Write the equation of the chord
For a parabola \( y^2 = 4ax \), the equation of the chord with midpoint \( ( \alpha, \beta ) \) can be expressed as:
\[ y - \beta = m(x - \alpha) \]
However, we can also use the formula derived from the parabola's properties:
\[ y \beta - 2ax + \alpha = \beta^2 - 4a\alpha \]

### Step 3: Substitute the fixed point into the chord equation
Since the chord passes through the fixed point \( (h, k) \), we substitute \( x = h \) and \( y = k \) into the chord equation:
\[ k \beta - 2ah + \alpha = \beta^2 - 4a\alpha \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ k \beta - 2ah + \alpha + 4a\alpha - \beta^2 = 0 \]
This simplifies to:
\[ k \beta - \beta^2 + (1 + 4a)\alpha - 2ah = 0 \]

### Step 5: Express in terms of \( \alpha \) and \( \beta \)
Rearranging further, we can express this as:
\[ \beta^2 - k \beta + (2ah - (1 + 4a)\alpha) = 0 \]

### Step 6: Identify the locus
For this quadratic equation in \( \beta \) to have real solutions, the discriminant must be non-negative:
\[ D = k^2 - 4(2ah - (1 + 4a)\alpha) \geq 0 \]
This gives us the condition for the locus of midpoints \( ( \alpha, \beta ) \).

### Step 7: Final equation of the locus
Rearranging the discriminant condition leads to:
\[ k^2 - 8ah + 4(1 + 4a)\alpha \geq 0 \]
This can be rearranged to find the relationship between \( \alpha \) and \( \beta \).

### Final Result
The locus of the midpoints of the chords of the parabola passing through the point \( (h, k) \) is given by the equation:
\[ \beta^2 - k\beta + 2ah - (1 + 4a)\alpha = 0 \]