If the foci of `(x^(2))/(16)+(y^(2))/(4)=1` and `(x^(2))/(a^(2))-(y^(2))/(3)=1` coincide, the value of a is

Options

A. 3
B. 2
C. `(1)/(sqrt(3))`
D. `sqrt(3)`

Explanation

To solve the problem, we need to find the value of \( a \) such that the foci of the given ellipse and hyperbola coincide.

### Step-by-Step Solution:

1. **Identify the given equations**:
- The equation of the ellipse is \(\frac{x^2}{16} + \frac{y^2}{4} = 1\).
- The equation of the hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{3} = 1\).

2. **Find the foci of the ellipse**:
- For an ellipse of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the foci are located at \((\pm ae, 0)\), where \(e = \sqrt{1 - \frac{b^2}{a^2}}\).
- Here, \(a^2 = 16\) and \(b^2 = 4\).
- Thus, \(a = 4\) and \(b = 2\).
- Calculate \(e\):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]
- Therefore, the foci of the ellipse are:
\[ (\pm ae, 0) = \left(\pm 4 \cdot \frac{\sqrt{3}}{2}, 0\right) = (\pm 2\sqrt{3}, 0) \]

3. **Find the foci of the hyperbola**:
- For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci are located at \((\pm ae, 0)\), where \(e = \sqrt{1 + \frac{b^2}{a^2}}\).
- Here, \(b^2 = 3\) and \(a^2 = a^2\).
- Calculate \(e\):
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{a^2}} \]
- Therefore, the foci of the hyperbola are:
\[ (\pm ae, 0) = \left(\pm a \sqrt{1 + \frac{3}{a^2}}, 0\right) \]

4. **Set the foci equal**:
- Since the foci of the ellipse and hyperbola coincide, we set:
\[ 2\sqrt{3} = a \sqrt{1 + \frac{3}{a^2}} \]

5. **Square both sides**:
\[ (2\sqrt{3})^2 = a^2 \left(1 + \frac{3}{a^2}\right) \]
\[ 12 = a^2 + 3 \]

6. **Solve for \(a^2\)**:
\[ a^2 = 12 - 3 = 9 \]
\[ a = \sqrt{9} = 3 \]

### Final Answer:
The value of \( a \) is \( 3 \).

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