P is parabola `y^2=4ax` and locus of mid points of all chords of this parabola passing through the vertex of P is a parabola P'. Axis of parabola P' will be

Options

A. parallel to axis P but dstinct
B. same as the axis of P
C. perpendicular to axis of P
D. None of these

Answer

To solve the problem, we need to find the locus of the midpoints of all chords of the parabola \( y^2 = 4ax \) that pass through the vertex (origin) of the parabola. Let's go through the solution step by step.

### Step 1: Understand the parabola and its properties
The given parabola is \( y^2 = 4ax \), which opens to the right. The vertex of this parabola is at the origin (0, 0).

### Step 2: Identify the endpoints of the chord
Let the endpoints of the chord be \( O(0, 0) \) (the vertex) and \( C(at^2, 2at) \), where \( t \) is a parameter. The point \( C \) lies on the parabola.

### Step 3: Find the midpoint of the chord
The coordinates of the midpoint \( M \) of the chord \( OC \) can be calculated as follows:
\[ M(h, k) = \left( \frac{0 + at^2}{2}, \frac{0 + 2at}{2} \right) = \left( \frac{at^2}{2}, at \right) \]
Thus, the coordinates of the midpoint \( M \) are \( \left( \frac{at^2}{2}, at \right) \).

### Step 4: Eliminate the parameter \( t \)
To find the locus of the midpoints, we need to eliminate the parameter \( t \) from the coordinates of \( M \):
- From \( k = at \), we can express \( t \) as \( t = \frac{k}{a} \).
- Substitute \( t \) into the expression 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 5: Rearranging the equation
Now, rearranging the equation \( h = \frac{k^2}{2a} \) gives us:
\[ k^2 = 2ah \]
This is the equation of a parabola.

### Step 6: Identify the form of the locus
Replacing \( h \) with \( x \) and \( k \) with \( y \), we get the equation of the locus:
\[ y^2 = 2ax \]
This represents a parabola \( P' \).

### Step 7: Determine the axis of the parabola \( P' \)
The parabola \( y^2 = 2ax \) opens to the right and has its axis along the x-axis, similar to the original parabola \( y^2 = 4ax \).

### Conclusion
The axis of the parabola \( P' \) is the x-axis.