If the ellipse `x^(2)+lambda^(2)y^(2)=lambda^(2)a^(2) , lambda^(2) gt1` is confocal with the hyperbola `x^(2)-y^(2)=a^(2)`, then

Options

A. ratio of eccentricities of ellipse and hyperbola is `1:sqrt(3)`
B. ratio of major axis of ellipse and transverse axis of hyperbola is `sqrt(3):1`
C. `The ellipse and hyperbola cuts each other orthogonally
D. ratio of length of latusrectumof ellipse and hyperbola is `1:3`

Answer

To solve the problem, we need to analyze the given ellipse and hyperbola, and determine their properties, particularly focusing on their foci and eccentricities.

### Step-by-Step Solution:

1. **Identify the given equations:**
- The ellipse is given by the equation:
\[ x^2 + \lambda^2 y^2 = \lambda^2 a^2 \]
- The hyperbola is given by the equation:
\[ x^2 - y^2 = a^2 \]

2. **Determine the foci of the ellipse:**
- The standard form of the ellipse is:
\[ \frac{x^2}{\lambda^2 a^2} + \frac{y^2}{a^2} = 1 \]
- The foci of the ellipse are located at:
\[ (\pm \lambda a, 0) \]
- Here, the eccentricity \(e_1\) of the ellipse is given by:
\[ e_1 = \sqrt{1 - \frac{a^2}{\lambda^2 a^2}} = \sqrt{1 - \frac{1}{\lambda^2}} = \frac{\sqrt{\lambda^2 - 1}}{\lambda} \]

3. **Determine the foci of the hyperbola:**
- The standard form of the hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1 \]
- The foci of the hyperbola are located at:
\[ (\pm \sqrt{a^2 + a^2}, 0) = (\pm a\sqrt{2}, 0) \]
- The eccentricity \(e_2\) of the hyperbola is given by:
\[ e_2 = \sqrt{1 + \frac{a^2}{a^2}} = \sqrt{2} \]

4. **Set the foci equal for confocality:**
- Since the ellipse and hyperbola are confocal, we have:
\[ \lambda a = a\sqrt{2} \]
- Dividing both sides by \(a\) (assuming \(a \neq 0\)):
\[ \lambda = \sqrt{2} \]

5. **Calculate the eccentricities:**
- For the ellipse:
\[ e_1 = \frac{\sqrt{\lambda^2 - 1}}{\lambda} = \frac{\sqrt{2 - 1}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
- For the hyperbola:
\[ e_2 = \sqrt{2} \]

6. **Find the ratio of eccentricities:**
- The ratio of the eccentricities \( \frac{e_1}{e_2} \):
\[ \frac{e_1}{e_2} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2} \]

7. **Check the ratio of the major axes:**
- The major axis of the ellipse is \(2\lambda a = 2\sqrt{2}a\).
- The transverse axis of the hyperbola is \(2a\).
- The ratio of the major axes:
\[ \frac{2\sqrt{2}a}{2a} = \sqrt{2} \]

8. **Check the latus rectum:**
- The latus rectum of the ellipse is given by:
\[ \frac{2a^2}{\lambda} = \frac{2a^2}{\sqrt{2}} = \sqrt{2}a^2 \]
- The latus rectum of the hyperbola is given by:
\[ \frac{2a^2}{a} = 2a \]
- The ratio of the latus rectum:
\[ \frac{\sqrt{2}a^2}{2a} = \frac{\sqrt{2}}{2}a \]

### Final Answer:
- The options that are correct based on the calculations are:
- The ratio of eccentricities is \( \frac{1}{2} \).
- The ratio of the major axes is \( \sqrt{2} \).
- The latus rectum ratio is \( \frac{\sqrt{2}}{2} \).