If the foci of the conics `x^(2)/a^(2) + y^(2)/7 =1 and x^(2)/144 - y^(2)/81 = 1/25` where to coincide, then what is the value of `a` ?

Options

A. 2
B. 3
C. 4
D. 16

Answer

To find the value of \( a \) such that the foci of the given conics coincide, we will follow these steps:

### Step 1: Identify the equations of the conics
The equations given are:
1. Ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{7} = 1 \)
2. Hyperbola: \( \frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25} \)

### Step 2: Find the foci of the ellipse
For the ellipse, the foci can be calculated using the formula:
\[ c = ae \]
where \( e \) is the eccentricity given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Here, \( b^2 = 7 \), so:
\[ e = \sqrt{1 - \frac{7}{a^2}} \]
Thus, the foci of the ellipse are:
\[ (\pm ae, 0) = \left(\pm a \sqrt{1 - \frac{7}{a^2}}, 0\right) \]

### Step 3: Find the foci of the hyperbola
For the hyperbola, the foci can be calculated using the formula:
\[ c = ae \]
where \( e \) for the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Here, \( b^2 = 81 \) and \( a^2 = 144 \), so:
\[ e = \sqrt{1 + \frac{81}{144}} = \sqrt{\frac{225}{144}} = \frac{15}{12} = \frac{5}{4} \]
Thus, the foci of the hyperbola are:
\[ (\pm ae, 0) = \left(\pm \frac{144}{25} \cdot \frac{5}{4}, 0\right) = \left(\pm \frac{180}{25}, 0\right) = \left(\pm 7.2, 0\right) \]

### Step 4: Set the foci equal to each other
Since the foci of both conics coincide, we set:
\[ a \sqrt{1 - \frac{7}{a^2}} = 7.2 \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ a^2 \left(1 - \frac{7}{a^2}\right) = 7.2^2 \]
This simplifies to:
\[ a^2 - 7 = 51.84 \]
Thus:
\[ a^2 = 51.84 + 7 = 58.84 \]

### Step 6: Solve for \( a \)
Taking the square root:
\[ a = \sqrt{58.84} \approx 7.67 \]

### Step 7: Check for possible values
Since the problem asks for the value of \( a \) and we are looking for a positive value, we conclude:
\[ a \approx 7.67 \]

### Final Answer
The value of \( a \) is approximately \( 7.67 \).