Find the locus of the middle points of all chords of the parabola `y^(2) = 4ax`, which are drawn through the vertex.

Options

A. `y^(2) = -4ax`
B. `y^(2) = 2ax`
C. `y^(2) = 4ax`
D. `y^(2) = -2ax`

Answer

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

### Step 1: Identify the vertex and the parabola
The vertex of the parabola \( y^2 = 4ax \) is at the point \( (0, 0) \).

### Step 2: Define the endpoints of the chord
Let one endpoint of the chord be at the vertex \( O(0, 0) \) and the other endpoint be at a point \( P(t) \) on the parabola. The coordinates of point \( P \) can be expressed in terms of a parameter \( t \) as:
\[ P(t) = (at^2, 2at) \]
where \( a \) is a constant related to the parabola.

### Step 3: Find the midpoint of the chord
The midpoint \( M(h, k) \) of the chord \( OP \) can be calculated using the midpoint formula:
\[ M(h, k) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of points \( O(0, 0) \) and \( P(at^2, 2at) \):
\[ M(h, k) = \left( \frac{0 + at^2}{2}, \frac{0 + 2at}{2} \right) = \left( \frac{at^2}{2}, at \right) \]
Thus, we have:
\[ h = \frac{at^2}{2} \quad \text{and} \quad k = at \]

### Step 4: Express \( t \) in terms of \( k \)
From the equation \( k = at \), we can express \( t \) as:
\[ t = \frac{k}{a} \]

### Step 5: Substitute \( t \) back into the equation for \( h \)
Now substitute \( t \) into the equation for \( h \):
\[ h = \frac{a\left(\frac{k}{a}\right)^2}{2} = \frac{a \cdot \frac{k^2}{a^2}}{2} = \frac{k^2}{2a} \]

### Step 6: Rearranging the equation
Rearranging the equation \( h = \frac{k^2}{2a} \) gives:
\[ k^2 = 2ah \]

### Step 7: Write the locus equation
Replacing \( h \) with \( x \) and \( k \) with \( y \), we obtain the locus of the midpoints:
\[ y^2 = 2ax \]

### Conclusion
The locus of the midpoints of all chords of the parabola \( y^2 = 4ax \) that pass through the vertex is given by the equation:
\[ y^2 = 2ax \]