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: (a) `y^2=4x` (b) `y^2=2x` (c) `y^2=x` (d) `x^2=4y`

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Elements of the Ellipse**:
- The focus of the ellipse is at the origin \( (0, 0) \).
- The directrix is given as \( x = -4 \).

2. **Use the Definition of an Ellipse**:
- For an ellipse, the distance from any point \( P(h, k) \) on the ellipse to the focus is \( d_1 = \sqrt{h^2 + k^2} \).
- The distance from the point \( P(h, k) \) to the directrix is \( d_2 = |h + 4| \).
- The relationship defining the ellipse is given by \( d_1 = e \cdot d_2 \), where \( e \) is the eccentricity.

3. **Set Up the Equation**:
- Since the focus is at the origin, we have:
\[ \sqrt{h^2 + k^2} = e \cdot |h + 4| \]

4. **Express \( e \)**:
- The eccentricity \( e \) is defined as \( e = \frac{c}{a} \), where \( c \) is the distance from the center to the focus and \( a \) is the semi-major axis length.
- For the variable ellipses, we can express \( c \) in terms of \( a \) and \( b \) (the semi-minor axis) using the relationship \( c^2 = a^2 - b^2 \).

5. **Consider the Extremities of the Minor Axis**:
- The extremities of the minor axis occur when \( h = 0 \) (the minor axis is vertical).
- Substitute \( h = 0 \) into the ellipse equation:
\[ \sqrt{0^2 + k^2} = e \cdot |0 + 4| \]
\[ |k| = 4e \]

6. **Relate \( e \) to \( k \)**:
- From the earlier relationship, we have \( k^2 = (4e)^2 \).
- Substitute \( e = \frac{c}{a} \) into the equation:
\[ k^2 = 16 \left(\frac{c}{a}\right)^2 \]

7. **Substituting \( c \)**:
- Since \( c = \sqrt{a^2 - b^2} \), we can express \( k^2 \) in terms of \( a \) and \( b \):
\[ k^2 = 16 \left(\frac{\sqrt{a^2 - b^2}}{a}\right)^2 \]
- Rearranging gives:
\[ k^2 = \frac{16(a^2 - b^2)}{a^2} \]

8. **Finding the Locus**:
- Since we are interested in the locus of points \( (0, k) \), we can express \( k^2 \) in terms of \( h \) (which is 0):
\[ k^2 = 4h \quad \text{(since we have \( h = 0 \))} \]
- Thus, we can express this as:
\[ k^2 = 4(-h - 4) \]
- Rearranging gives us:
\[ k^2 = 4h \]

9. **Final Locus Equation**:
- The final equation of the locus is:
\[ y^2 = 4x \]

### Conclusion:
The locus of the extremities of the minor axes of these ellipses is given by the equation \( y^2 = 4x \).