In an ellipse and a hyperbola having centre at origin and foci on `(pm 10, 0)`, if the distance between their Directix (both in positive direction of x axis) is 10 units, find Difference of Sqaures of the Lengths of their major Axis and Transverse axis

Options

A. 400
B. 300
C. 200
D. 160

Explanation

To solve the problem, we need to find the difference of squares of the lengths of the major axis of an ellipse and the transverse axis of a hyperbola, given certain conditions about their foci and directrices.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The foci of both the ellipse and hyperbola are at \((\pm 10, 0)\).
- The distance between their directrices in the positive direction of the x-axis is 10 units.

2. **Define the Parameters:**
- For the ellipse:
- Let \( a_1 \) be the semi-major axis.
- Let \( e_1 \) be the eccentricity.
- For the hyperbola:
- Let \( a_2 \) be the semi-transverse axis.
- Let \( e_2 \) be the eccentricity.

3. **Relate the Foci to the Semi-Axes:**
- For the ellipse:
\[ e_1 = \frac{c_1}{a_1} \quad \text{where } c_1 = 10 \]
Thus,
\[ e_1 = \frac{10}{a_1} \]
- For the hyperbola:
\[ e_2 = \frac{c_2}{a_2} \quad \text{where } c_2 = 10 \]
Thus,
\[ e_2 = \frac{10}{a_2} \]

4. **Use the Directrix Formula:**
- The directrix \( D \) for the ellipse is given by \( \frac{a_1}{e_1} \) and for the hyperbola is \( \frac{a_2}{e_2} \).
- The distance between the directrices is:
\[ \frac{a_1}{e_1} - \frac{a_2}{e_2} = 10 \]

5. **Substituting the Eccentricities:**
- Substitute \( e_1 \) and \( e_2 \) into the directrix equation:
\[ \frac{a_1}{\frac{10}{a_1}} - \frac{a_2}{\frac{10}{a_2}} = 10 \]
- This simplifies to:
\[ \frac{a_1^2}{10} - \frac{a_2^2}{10} = 10 \]
- Multiplying through by 10 gives:
\[ a_1^2 - a_2^2 = 100 \]

6. **Finding the Difference of Squares of Lengths:**
- The length of the major axis of the ellipse is \( 2a_1 \) and the length of the transverse axis of the hyperbola is \( 2a_2 \).
- The difference of squares is:
\[ (2a_1)^2 - (2a_2)^2 = 4(a_1^2 - a_2^2) \]
- Substitute \( a_1^2 - a_2^2 = 100 \):
\[ 4(100) = 400 \]

7. **Final Answer:**
- The difference of squares of the lengths of their major axis and transverse axis is \( \boxed{400} \).