The locus of the middle points of the chords of the parabola `y^(2)=4ax` which pass through the focus, is

Options

A. `y^(2)+2ax+2a^(2)=0`
B. `y^(2)-ax+2a^(2)=0`
C. `y^(2)-2ax+2a^(2)=0`
D. `y^(2)-2ax+a^(2)=0`

Explanation

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

### Step 1: Understand the Parabola and its Focus
The given parabola is \( y^2 = 4ax \). The focus of this parabola is the point \( (a, 0) \).

### Step 2: Consider the Midpoint of the Chord
Let the midpoint of the chord be \( (h, k) \). The equation of the chord can be expressed using the midpoint formula. For a chord with midpoint \( (h, k) \), the equation can be derived from the general form of the parabola.

### Step 3: Use the Chord Equation
The equation of the chord passing through the focus can be expressed as:
\[ T = S_1 \]
where \( T \) is the equation of the chord and \( S_1 \) is the value of the parabola at the midpoint.

### Step 4: Calculate \( T \)
Substituting \( (h, k) \) into the parabola equation:
\[ k \cdot k = 4a \cdot h \implies k^2 = 4ah \]
This gives us the equation of the chord in terms of \( h \) and \( k \).

### Step 5: Calculate \( S_1 \)
For the point \( (h, k) \):
\[ S_1 = k^2 - 4ah \]
Now we set \( T = S_1 \):
\[ k \cdot k - 2a(h) + h = k^2 - 4ah \]

### Step 6: Substitute the Focus Coordinates
Since the chord passes through the focus \( (a, 0) \), we substitute \( x = a \) and \( y = 0 \) into the chord equation:
\[ 0 \cdot k - 2a + h = k^2 - 4ah \]
This simplifies to:
\[ -2a + h = k^2 - 4ah \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ k^2 - 4ah + h + 2a = 0 \]

### Step 8: Replace Variables
Replacing \( h \) with \( x \) and \( k \) with \( y \):
\[ y^2 - 4ax + x + 2a = 0 \]

### Step 9: Final Rearrangement
Bringing all terms to one side:
\[ y^2 - 4ax + x + 2a = 0 \]
This can be rearranged to:
\[ y^2 - 2ax + 2a = 0 \]

### Conclusion
Thus, the locus of the midpoints of the chords of the parabola \( y^2 = 4ax \) that pass through the focus is given by:
\[ y^2 - 2ax + 2a = 0 \]