The locus of the mid-point of the line segment joning the focus to a moving point on the parabola `y^(2)=4ax` is another parabola with directrix

Options

A. `x= -a`
B. `x = -a//2`
C. `x=0`
D. `x=a//2`

Answer

To find the locus of the midpoint of the line segment joining the focus to a moving point on the parabola \(y^2 = 4ax\), we can follow these steps:

### Step 1: Identify the focus and a point on the parabola
The given parabola is \(y^2 = 4ax\). The focus of this parabola is at the point \(F(a, 0)\). Let the moving point on the parabola be \(P(x_1, y_1)\). Since \(P\) lies on the parabola, it satisfies the equation:
\[ y_1^2 = 4ax_1 \]

### Step 2: Find the midpoint of the segment joining the focus and the point on the parabola
The midpoint \(M(h, k)\) of the segment joining the focus \(F(a, 0)\) and the point \(P(x_1, y_1)\) can be calculated using the midpoint formula:
\[ h = \frac{x_1 + a}{2}, \quad k = \frac{y_1 + 0}{2} = \frac{y_1}{2} \]

### Step 3: Express \(x_1\) and \(y_1\) in terms of \(h\) and \(k\)
From the expression for \(h\):
\[ x_1 = 2h - a \]
Substituting this into the parabola equation \(y_1^2 = 4ax_1\):
\[ y_1^2 = 4a(2h - a) = 8ah - 4a^2 \]

### Step 4: Substitute \(y_1\) in terms of \(k\)
Since \(k = \frac{y_1}{2}\), we have:
\[ y_1 = 2k \]
Substituting \(y_1\) into the equation gives:
\[ (2k)^2 = 8ah - 4a^2 \]
This simplifies to:
\[ 4k^2 = 8ah - 4a^2 \]
Dividing through by 4:
\[ k^2 = 2ah - a^2 \]

### Step 5: Rearranging to find the locus
Rearranging the equation gives:
\[ k^2 = 2a(h - \frac{a}{2}) \]
This is the equation of a parabola in the form \(k^2 = 4p(h - h_0)\), where \(p = \frac{a}{2}\) and \(h_0 = \frac{a}{2}\).

### Conclusion
The locus of the midpoint \(M(h, k)\) is a parabola with the directrix \(h = \frac{a}{2}\).