The locus of the midpoints of the focal chords of the parabola `y^(2)=4ax` is

Options

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

Answer

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

### Step 1: Identify the endpoints of the focal chord
For the parabola \( y^2 = 4ax \), the endpoints of a focal chord can be expressed in parametric form as:
- Point 1: \( (at^2, 2at) \)
- Point 2: \( \left(\frac{a}{t^2}, -\frac{2a}{t}\right) \)

### Step 2: Find the midpoint of the focal chord
The midpoint \( (h, k) \) of the focal chord can be calculated using the coordinates of the endpoints:
\[ h = \frac{at^2 + \frac{a}{t^2}}{2}, \quad k = \frac{2at - \frac{2a}{t}}{2} \]

### Step 3: Simplify the expressions for \( h \) and \( k \)
1. For \( h \):
\[ h = \frac{a}{2} \left(t^2 + \frac{1}{t^2}\right) \]
2. For \( k \):
\[ k = a \left(t - \frac{1}{t}\right) \]

### Step 4: Relate \( h \) and \( k \)
To find a relationship between \( h \) and \( k \), we can express \( t \) in terms of \( k \):
\[ k = a \left(t - \frac{1}{t}\right) \implies \frac{k}{a} = t - \frac{1}{t} \]
Let \( u = t - \frac{1}{t} \). Then, we can express \( t^2 + \frac{1}{t^2} \) in terms of \( u \):
\[ t^2 + \frac{1}{t^2} = u^2 + 2 \]

### Step 5: Substitute back into the equation for \( h \)
Substituting \( u \) back into the equation for \( h \):
\[ h = \frac{a}{2} \left(u^2 + 2\right) = \frac{a}{2} \left(\left(\frac{k}{a}\right)^2 + 2\right) \]
This simplifies to:
\[ h = \frac{a}{2} \left(\frac{k^2}{a^2} + 2\right) \]

### Step 6: Rearranging the equation
Multiplying through by \( 2a^2 \) gives:
\[ 2ah = k^2 + 4a^2 \]
Rearranging this, we have:
\[ k^2 = 2ah - 4a^2 \]

### Step 7: Final form of the locus
This can be rewritten as:
\[ k^2 = 2a(h - 2a) \]
Substituting \( h \) with \( x \) and \( k \) with \( y \), we get the locus of the midpoints:
\[ y^2 = 2a(x - 2a) \]

### Conclusion
The locus of the midpoints of the focal chords of the parabola \( y^2 = 4ax \) is given by the equation:
\[ y^2 = 2a(x - 2a) \]