If `P(alpha, beta)`, the point of intersection of the ellipse `x^2/a^2+y^2/a^2 (1-e^2)=1` and hyperbola `x^2/a^2-y^2/(a^2(E^2-1))=1/4`is equidistant from the foci of the curves all lying in the right of y-axis then

Options

A. `2 alpha =a (2e +E)`
B. `a- ealpha = E alpha -alpha//2`
C. `E =(sqrt(e^(2)+24)-3e)/(2)`
D. `E=(sqrt(e^(2)+12)-3e)/(2)`

Explanation

To solve the problem, we need to analyze the given equations of the ellipse and hyperbola, find their point of intersection, and then determine the conditions under which this point is equidistant from the foci of both curves.

### Step-by-Step Solution

1. **Identify the equations of the curves:**
- The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{a^2(1-e^2)} = 1 \]
- The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{a^2(e^2-1)} = \frac{1}{4} \]

2. **Rewrite the hyperbola equation:**
- We can rewrite the hyperbola equation by multiplying through by 4:
\[ \frac{4x^2}{a^2} - \frac{4y^2}{a^2(e^2-1)} = 1 \]
- This implies:
\[ \frac{x^2}{\frac{a^2}{4}} - \frac{y^2}{\frac{a^2(e^2-1)}{4}} = 1 \]
- Here, we can identify \( a' = \frac{a}{2} \) and \( b' = \frac{a\sqrt{(e^2-1)}}{2} \).

3. **Find the foci of the ellipse and hyperbola:**
- The foci of the ellipse are located at \( (0, \pm ae) \).
- The foci of the hyperbola are located at \( (\pm a', 0) = \left(\pm \frac{a}{2}, 0\right) \).

4. **Set up the condition for equidistance:**
- Let \( P(\alpha, \beta) \) be the point of intersection. The distances from \( P \) to the foci of the ellipse and hyperbola must be equal.
- For the ellipse, the distance to the focus \( (0, ae) \) is:
\[ d_1 = \sqrt{(\alpha - 0)^2 + (\beta - ae)^2} = \sqrt{\alpha^2 + (\beta - ae)^2} \]
- For the hyperbola, the distance to the focus \( \left(\frac{a}{2}, 0\right) \) is:
\[ d_2 = \sqrt{\left(\alpha - \frac{a}{2}\right)^2 + \beta^2} \]

5. **Set the distances equal:**
- We set \( d_1 = d_2 \):
\[ \sqrt{\alpha^2 + (\beta - ae)^2} = \sqrt{\left(\alpha - \frac{a}{2}\right)^2 + \beta^2} \]
- Squaring both sides gives:
\[ \alpha^2 + (\beta - ae)^2 = \left(\alpha - \frac{a}{2}\right)^2 + \beta^2 \]

6. **Expand and simplify:**
- Expanding both sides:
\[ \alpha^2 + \beta^2 - 2\beta ae + a^2e^2 = \alpha^2 - a\alpha + \frac{a^2}{4} + \beta^2 \]
- Canceling \( \alpha^2 \) and \( \beta^2 \) from both sides:
\[ -2\beta ae + a^2e^2 = -a\alpha + \frac{a^2}{4} \]
- Rearranging gives:
\[ 2\beta ae - a\alpha = \frac{a^2}{4} - a^2e^2 \]

7. **Solve for \(\alpha\) and \(\beta\):**
- Rearranging the equation yields:
\[ a\alpha = 2\beta ae + a^2(1/4 - e^2) \]
- Thus:
\[ \alpha = 2\beta e + \frac{a}{4} - ae^2 \]

8. **Conclusion:**
- The relationship between \(\alpha\) and \(\beta\) can be further analyzed based on the specific values of \(a\) and \(e\) to determine the correct option from the given choices.