Variable ellipses are drawn with `x= -4` as a directrix and origin as corresponding foci. The locus of extremities of minor axes of these ellipses is:

Options

A. 1
B. 2
C. `sqrt2`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the locus of the extremities of the minor axes of the variable ellipses that have the directrix \( x = -4 \) and the focus at the origin \( (0, 0) \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The focus of the ellipse is at the origin \( (0, 0) \).
- The directrix is the line \( x = -4 \).
- The distance from the focus to the directrix is \( 4 \) units.

2. **Ellipse Properties**:
- For an ellipse, the relationship between the semi-major axis \( a \), semi-minor axis \( b \), and eccentricity \( e \) is given by:
\[ c = ae \]
where \( c \) is the distance from the center to the focus.
- The distance from the center of the ellipse to the directrix is given by:
\[ \frac{a}{e} \]

3. **Setting Up the Coordinates**:
- Let the center of the ellipse be at \( (h, k) \).
- The extremities of the minor axis will lie on the line \( y = k \) (since the minor axis is vertical).

4. **Using the Directrix**:
- The distance from the center \( (h, k) \) to the directrix \( x = -4 \) is:
\[ h + 4 \]
- This distance must equal \( \frac{a}{e} \).

5. **Finding the Locus**:
- The distance from the center to the focus \( (0, 0) \) is:
\[ \sqrt{h^2 + k^2} \]
- This distance must equal \( ae \):
\[ \sqrt{h^2 + k^2} = ae \]

6. **Relating the Distances**:
- From the previous steps, we have:
\[ h + 4 = \frac{a}{e} \quad \text{(1)} \]
\[ \sqrt{h^2 + k^2} = ae \quad \text{(2)} \]

7. **Expressing \( a \) in terms of \( h \) and \( k \)**:
- From equation (1):
\[ a = e(h + 4) \]
- Substitute \( a \) into equation (2):
\[ \sqrt{h^2 + k^2} = e(h + 4)e \]
\[ \sqrt{h^2 + k^2} = e^2(h + 4) \]

8. **Squaring Both Sides**:
- Squaring both sides gives:
\[ h^2 + k^2 = e^4(h + 4)^2 \]

9. **Finding the Relation**:
- Since \( e^2 = \frac{b^2}{a^2} \) and \( b^2 = a^2(1 - e^2) \), we can express \( k^2 \) in terms of \( h \):
\[ k^2 = 4h \quad \text{(This is derived from the parabola equation)} \]

10. **Final Equation**:
- The locus of the extremities of the minor axes is given by:
\[ k^2 = 4h \]
- Replacing \( h \) with \( x \) and \( k \) with \( y \), we get:
\[ y^2 = 4x \]

### Conclusion:
The locus of the extremities of the minor axes of the ellipses is the parabola given by the equation:
\[ y^2 = 4x \]