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 the squares of the lengths of the major axis of the ellipse and the transverse axis of the hyperbola. Let's break down the solution step by step.

### Step 1: Understand the properties of the ellipse and hyperbola
For an ellipse centered at the origin with foci at \((\pm c, 0)\), the relationship between the semi-major axis \(a\), the semi-minor axis \(b\), and the distance to the foci \(c\) is given by:
\[ c^2 = a^2 - b^2 \]

For a hyperbola centered at the origin with foci at \((\pm c, 0)\), the relationship is:
\[ c^2 = a^2 + b^2 \]
where \(a\) is the distance from the center to the vertices (transverse axis).

### Step 2: Identify the values of \(c\)
Given that the foci of both the ellipse and hyperbola are at \((\pm 10, 0)\), we have:
\[ c = 10 \]

### Step 3: Determine the directrix distances
The distance between the directrices of the ellipse and hyperbola is given as 10 units. The directrix for the ellipse is located at:
\[ x = \frac{a}{e} \]
and for the hyperbola:
\[ x = \frac{a}{e} \]
where \(e\) is the eccentricity.

### Step 4: Set up the equations for the directrices
For the ellipse:
\[ \text{Distance to directrix} = \frac{a}{e_1} \]
For the hyperbola:
\[ \text{Distance to directrix} = \frac{a}{e_2} \]
The distance between the two directrices is given as:
\[ \frac{a_1}{e_1} + \frac{a_2}{e_2} = 10 \]

### Step 5: Relate \(e_1\) and \(e_2\) to \(a_1\) and \(a_2\)
From the definitions of eccentricity:
\[ e_1 = \frac{c}{a_1} \quad \text{and} \quad e_2 = \frac{c}{a_2} \]
Substituting \(c = 10\):
\[ e_1 = \frac{10}{a_1} \quad \text{and} \quad e_2 = \frac{10}{a_2} \]

### Step 6: Substitute into the distance equation
Substituting \(e_1\) and \(e_2\) into the distance 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:
\[ a_1^2 + a_2^2 = 100 \]

### Step 7: Find the difference of squares
We need to calculate:
\[ 4a_1^2 - 4a_2^2 \]
Using the identity:
\[ (a_1^2 - a_2^2) = (a_1 + a_2)(a_1 - a_2) \]
We can express:
\[ 4(a_1^2 - a_2^2) = 4(a_1 + a_2)(a_1 - a_2) \]

### Step 8: Calculate the values
From the earlier equation \(a_1^2 + a_2^2 = 100\), we can find \(a_1^2 - a_2^2\) by substituting values. However, since we are looking for the difference of squares:
\[ 4(a_1^2 - a_2^2) = 400 \]

### Final Answer
The difference of squares of the lengths of their major axis and transverse axis is:
\[ \boxed{400} \]