The locus of mid - points of all chords of parabola `y^(2)=4x,` for which all cirlces drawn taking them as diameters passes through the vertex of the parabola is a conic whose length of the smallest focal chord is equal to

Options

A. 1 units
B. 2 units
C. 3 units
D. 4 units

Answer

To solve the problem, we need to find the length of the smallest focal chord of the parabola given by the equation \( y^2 = 4x \). We will follow these steps:

### Step 1: Understand the Parabola
The given parabola is \( y^2 = 4x \). This is a standard form of a parabola that opens to the right with its vertex at the origin (0,0).

### Step 2: General Point on the Parabola
For the parabola \( y^2 = 4x \), we can express a general point on the parabola in terms of a parameter \( t \):
\[ P(t) = (t^2, 2t) \]
where \( t \) is a parameter.

### Step 3: Chords Passing Through the Vertex
Let’s consider two points on the parabola, \( A(t_1) = (t_1^2, 2t_1) \) and \( B(t_2) = (t_2^2, 2t_2) \). The chord \( AB \) passes through the vertex (0,0).

### Step 4: Condition for Chords
For the chord \( AB \) to pass through the vertex, the slopes of the lines from the vertex to points \( A \) and \( B \) must satisfy the condition for perpendicularity:
\[ \text{slope of } AV \cdot \text{slope of } BV = -1 \]
Calculating the slopes:
\[ \text{slope of } AV = \frac{2t_1 - 0}{t_1^2 - 0} = \frac{2t_1}{t_1^2} \]
\[ \text{slope of } BV = \frac{2t_2 - 0}{t_2^2 - 0} = \frac{2t_2}{t_2^2} \]
Setting the product of slopes to -1:
\[ \frac{2t_1}{t_1^2} \cdot \frac{2t_2}{t_2^2} = -1 \]
This simplifies to:
\[ \frac{4t_1t_2}{t_1^2t_2^2} = -1 \implies t_1t_2 = -4 \]

### Step 5: Midpoint of the Chord
The midpoint \( M \) of the chord \( AB \) can be calculated as:
\[ H = \frac{t_1^2 + t_2^2}{2}, \quad K = \frac{2t_1 + 2t_2}{2} = t_1 + t_2 \]

Using the identity \( (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2 \):
\[ K^2 = H + 2(-4) \implies K^2 = H - 8 \]

### Step 6: Locus of Midpoints
Replacing \( H \) and \( K \) with \( x \) and \( y \) respectively, we have:
\[ y^2 = x - 8 \]
This represents a parabola that opens to the right with vertex at (8, 0).

### Step 7: Length of the Smallest Focal Chord
The length of the latus rectum (which is the smallest focal chord) of the parabola \( y^2 = 4ax \) is given by \( 4a \). For our parabola \( y^2 = 4x \), we have \( a = 1 \):
\[ \text{Length of the latus rectum} = 4 \times 1 = 4 \]

### Conclusion
Thus, the length of the smallest focal chord is \( \boxed{4} \).