If A,A' are the vertices S,S' are the foci and Z,Z' are the feet of the directrices of an ellipse with centre C, then CS CA ,CZ are in

Options

A. A.P
B. G.P
C. H.P
D. none of these

Answer

To solve the problem, we will analyze the relationship between the distances \( CS \), \( CA \), and \( CZ \) for an ellipse centered at point \( C \).

### Step-by-Step Solution:

1. **Define the Ellipse**:
We consider the standard form of the ellipse given by the equation:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. The center \( C \) of the ellipse is at the origin \((0,0)\).

2. **Identify Key Points**:
- The vertices \( A \) and \( A' \) are located at \( (a, 0) \) and \( (-a, 0) \) respectively.
- The foci \( S \) and \( S' \) are located at \( (ae, 0) \) and \( (-ae, 0) \) respectively, where \( e = \sqrt{1 - \frac{b^2}{a^2}} \) is the eccentricity of the ellipse.
- The feet of the directrices \( Z \) and \( Z' \) are located at \( \left(\frac{a}{e}, 0\right) \) and \( \left(-\frac{a}{e}, 0\right) \) respectively.

3. **Calculate Distances**:
- The distance \( CS \) from the center \( C \) to the focus \( S \):
\[ CS = ae \]
- The distance \( CA \) from the center \( C \) to the vertex \( A \):
\[ CA = a \]
- The distance \( CZ \) from the center \( C \) to the foot of the directrix \( Z \):
\[ CZ = \frac{a}{e} \]

4. **Establish the Relationship**:
We need to show that \( CS \), \( CA \), and \( CZ \) are in geometric progression. For three quantities \( x \), \( y \), and \( z \) to be in geometric progression, the following condition must hold:
\[ y^2 = x \cdot z \]
Substituting our distances:
\[ CA^2 = CS \cdot CZ \]
This translates to:
\[ a^2 = (ae) \left(\frac{a}{e}\right) \]

5. **Verify the Equation**:
Simplifying the right-hand side:
\[ (ae) \left(\frac{a}{e}\right) = a^2 \]
Thus, we have:
\[ a^2 = a^2 \]
This confirms that the relationship holds true.

6. **Conclusion**:
Since \( CS \), \( CA \), and \( CZ \) satisfy the condition for being in geometric progression, we conclude that:
\[ CS, CA, CZ \text{ are in GP.} \]

### Final Answer:
The distances \( CS \), \( CA \), and \( CZ \) are in geometric progression (GP).

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