The eccentricities of the ellipse `x^(2)/alpha^(2)+y^(2)/beta^(2)=1,alphagtbeta` and `x^(2)/9+y^(2)/16=1` are equal. Which one of the following is correct?

Options

A. `4alpha = 3beta`
B. `alphabeta = 12`
C. `4beta = 3alpha`
D. `9alpha = 16beta`

Answer

To solve the problem, we need to find the relationship between the parameters of the two ellipses given that their eccentricities are equal.

### Step 1: Identify the parameters of the first ellipse
The first ellipse is given by the equation:
\[ \frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1 \]
Here, we identify:
- \( a = \alpha \) (semi-major axis)
- \( b = \beta \) (semi-minor axis)

### Step 2: Calculate the eccentricity of the first ellipse
The eccentricity \( e_1 \) of an ellipse is given by the formula:
\[ e = \frac{c}{a} \]
where \( c = \sqrt{a^2 - b^2} \).

For the first ellipse:
\[ c_1 = \sqrt{\alpha^2 - \beta^2} \]
Thus, the eccentricity \( e_1 \) becomes:
\[ e_1 = \frac{\sqrt{\alpha^2 - \beta^2}}{\alpha} \]

### Step 3: Identify the parameters of the second ellipse
The second ellipse is given by the equation:
\[ \frac{x^2}{9} + \frac{y^2}{16} = 1 \]
From this, we identify:
- \( a = 4 \) (since \( a^2 = 16 \))
- \( b = 3 \) (since \( b^2 = 9 \))

### Step 4: Calculate the eccentricity of the second ellipse
For the second ellipse:
\[ c_2 = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7} \]
Thus, the eccentricity \( e_2 \) becomes:
\[ e_2 = \frac{c_2}{a} = \frac{\sqrt{7}}{4} \]

### Step 5: Set the eccentricities equal
Since the eccentricities are given to be equal, we set \( e_1 = e_2 \):
\[ \frac{\sqrt{\alpha^2 - \beta^2}}{\alpha} = \frac{\sqrt{7}}{4} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{\alpha^2 - \beta^2}{\alpha^2} = \frac{7}{16} \]

### Step 7: Rearrange the equation
Rearranging the equation, we have:
\[ 16(\alpha^2 - \beta^2) = 7\alpha^2 \]
This simplifies to:
\[ 16\alpha^2 - 16\beta^2 = 7\alpha^2 \]
\[ 9\alpha^2 = 16\beta^2 \]

### Step 8: Express the relationship between alpha and beta
Dividing both sides by \( \alpha^2 \) gives:
\[ \frac{9}{16} = \frac{\beta^2}{\alpha^2} \]
Taking the square root of both sides results in:
\[ \frac{\beta}{\alpha} = \frac{3}{4} \]
Thus, we can express this as:
\[ 3\alpha = 4\beta \]

### Final Answer
The required relationship is:
\[ 3\alpha = 4\beta \]