The locus of the middle points of all chords of the parabola `y^(2)=4ax` passing through the vertex is

Options

A. a straight line
B. an ellipse
C. a parabola
D. a circle

Answer

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

### Step 1: Understand the Parabola
The given parabola is \( y^2 = 4ax \). The vertex of this parabola is at the origin \( (0, 0) \).

### Step 2: Parametric Form of the Parabola
The points on the parabola can be expressed in parametric form as:
\[ P(t) = (at^2, 2at) \]
where \( t \) is a parameter.

### Step 3: Identify the Chord
Let’s consider two points on the parabola \( P(t_1) \) and \( P(t_2) \) that form a chord passing through the vertex. The coordinates of these points are:
\[ P(t_1) = (at_1^2, 2at_1) \quad \text{and} \quad P(t_2) = (at_2^2, 2at_2) \]

### Step 4: Find the Midpoint of the Chord
The midpoint \( Q \) of the chord connecting \( P(t_1) \) and \( P(t_2) \) can be calculated using the midpoint formula:
\[ Q = \left( \frac{at_1^2 + at_2^2}{2}, \frac{2at_1 + 2at_2}{2} \right) = \left( \frac{a(t_1^2 + t_2^2)}{2}, a(t_1 + t_2) \right) \]

### Step 5: Express the Midpoint in Terms of \( h \) and \( k \)
Let the coordinates of the midpoint \( Q \) be \( (h, k) \). Thus, we have:
\[ h = \frac{a(t_1^2 + t_2^2)}{2} \quad \text{and} \quad k = a(t_1 + t_2) \]

### Step 6: Relate \( t_1 \) and \( t_2 \) to \( h \) and \( k \)
From the second equation, we can express \( t_1 + t_2 \) as:
\[ t_1 + t_2 = \frac{k}{a} \]
Now, we need to express \( t_1^2 + t_2^2 \) in terms of \( t_1 + t_2 \):
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2 = \left(\frac{k}{a}\right)^2 - 2t_1t_2 \]

### Step 7: Substitute Back to Find \( h \)
Now, substituting \( t_1^2 + t_2^2 \) back into the equation for \( h \):
\[ h = \frac{a}{2} \left( \left(\frac{k}{a}\right)^2 - 2t_1t_2 \right) \]
To find \( t_1t_2 \), we can use the fact that the product of the roots \( t_1 \) and \( t_2 \) can be expressed in terms of \( h \) and \( k \) as:
\[ t_1t_2 = \frac{h}{a} \]
Thus, substituting this into the equation for \( h \):
\[ h = \frac{a}{2} \left( \frac{k^2}{a^2} - 2 \cdot \frac{h}{a} \right) \]
This simplifies to:
\[ h = \frac{1}{2a}k^2 - h \]
Rearranging gives:
\[ 2h = \frac{1}{2a}k^2 \]
Multiplying through by \( 2a \):
\[ 4ah = k^2 \]

### Step 8: Final Equation of the Locus
Rearranging gives us the final equation of the locus:
\[ k^2 = 4ah \]
or
\[ y^2 = 4ax \]
This shows that the locus of the midpoints of all chords of the parabola that pass through the vertex is another parabola.

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