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 locus of the midpoints of all chords of the parabola \( y^2 = 4x \) such that the circles drawn with these chords as diameters pass through the vertex of the parabola.

### Step-by-Step Solution:

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

2. **Parameterization of the Parabola:**
We can parameterize points on the parabola using a parameter \( t \):
\[ P(t_1) = (at_1^2, 2at_1) = (t_1^2, 2t_1) \]
\[ P(t_2) = (at_2^2, 2at_2) = (t_2^2, 2t_2) \]
where \( a = 1 \) for our parabola.

3. **Finding the Midpoint of the Chord:**
The midpoint \( M \) of the chord joining points \( P(t_1) \) and \( P(t_2) \) is given by:
\[ M = \left( \frac{t_1^2 + t_2^2}{2}, \frac{2t_1 + 2t_2}{2} \right) = \left( \frac{t_1^2 + t_2^2}{2}, t_1 + t_2 \right) \]

4. **Condition for Circles to Pass through the Vertex:**
The circle with diameter endpoints \( P(t_1) \) and \( P(t_2) \) passes through the origin if the midpoint \( M \) satisfies the equation of the circle. The equation of the circle is:
\[ (x - M_x)^2 + (y - M_y)^2 = M_x^2 + M_y^2 \]
For the circle to pass through the origin, we need:
\[ M_x^2 + M_y^2 = \frac{(t_1^2 + t_2^2)^2}{4} + (t_1 + t_2)^2 \]

5. **Using the Identity:**
We can use the identity \( t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2 \) to simplify the expression:
\[ M_x^2 + M_y^2 = \frac{((t_1 + t_2)^2 - 2t_1t_2)^2}{4} + (t_1 + t_2)^2 \]

6. **Finding the Locus:**
Let \( s = t_1 + t_2 \) and \( p = t_1t_2 \). The locus of midpoints can be derived to yield a conic section. After simplification, we find:
\[ y^2 = 2x \]

7. **Finding the Length of the Smallest Focal Chord:**
The length of the smallest focal chord of the parabola \( y^2 = 4x \) is given by the formula:
\[ \text{Length} = 2a \]
where \( a = 1 \) in our case. Thus, the length of the smallest focal chord is:
\[ 2 \times 1 = 2 \]

### Final Answer:
The length of the smallest focal chord is \( 2 \).