Let O be the origin and A be a point on the curve `y^(2)=4x` then locus of the midpoint of OA is

Options

A. `x^(2)=4y`
B. `x^(2)=2y`
C. `y^(2)=16x`
D. `y^(2)=2x`

Answer

To find the locus of the midpoint of the line segment OA, where O is the origin (0, 0) and A is a point on the curve \(y^2 = 4x\), we can follow these steps:

### Step 1: Identify the coordinates of point A
Since point A lies on the curve \(y^2 = 4x\), we can express the coordinates of A in terms of a parameter. Let:
\[ A = (2h, 2k) \]
where \(k^2 = 2h\) (derived from the curve equation).

### Step 2: Verify the coordinates of A
Substituting \(A = (2h, 2k)\) into the curve equation:
\[ (2k)^2 = 4(2h) \]
This simplifies to:
\[ 4k^2 = 8h \]
Thus, we have:
\[ k^2 = 2h \]
This confirms that point A is indeed on the curve.

### Step 3: Find the midpoint M of OA
The coordinates of the midpoint M of the line segment OA can be calculated using the midpoint formula:
\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Here, \(O = (0, 0)\) and \(A = (2h, 2k)\). Therefore, we have:
\[ M = \left(\frac{0 + 2h}{2}, \frac{0 + 2k}{2}\right) = (h, k) \]

### Step 4: Express the relationship between h and k
From the earlier step, we know:
\[ k^2 = 2h \]
We can express this in terms of \(h\) and \(k\):
\[ h = \frac{k^2}{2} \]

### Step 5: Substitute h in terms of k into the midpoint coordinates
Now, substituting \(h\) in the coordinates of M:
\[ M = \left(\frac{k^2}{2}, k\right) \]

### Step 6: Find the locus of M
To find the locus, we eliminate the parameter \(h\) and express the relationship between \(x\) and \(y\). Let:
\[ x = \frac{k^2}{2} \quad \text{and} \quad y = k \]
Substituting \(k = y\) into the equation for \(x\):
\[ x = \frac{y^2}{2} \]
Rearranging gives us:
\[ y^2 = 2x \]

### Conclusion
Thus, the locus of the midpoint M of OA is given by the equation:
\[ y^2 = 2x \]