If the equation `x^2/(10-2a)+y^2/(4-2a)=1` represents an ellipse , then a lies in the interval

Options

A. `(-oo,5)`
B. `(2,5)`
C. `(-oo,2)`
D. `(5,oo)`

Answer

To determine the values of \( a \) for which the equation

\[ \frac{x^2}{10 - 2a} + \frac{y^2}{4 - 2a} = 1 \]

represents an ellipse, we need to ensure that both denominators \( 10 - 2a \) and \( 4 - 2a \) are positive.

### Step 1: Set up the inequalities

For the equation to represent an ellipse, we require:

1. \( 10 - 2a > 0 \)
2. \( 4 - 2a > 0 \)

### Step 2: Solve the first inequality

Starting with the first inequality:

\[ 10 - 2a > 0 \]

Rearranging gives:

\[ 10 > 2a \implies a < 5 \]

### Step 3: Solve the second inequality

Now, we solve the second inequality:

\[ 4 - 2a > 0 \]

Rearranging gives:

\[ 4 > 2a \implies a < 2 \]

### Step 4: Combine the results

From the two inequalities, we have:

1. \( a < 5 \)
2. \( a < 2 \)

The more restrictive condition is \( a < 2 \).

### Step 5: Determine the lower bound

Next, we need to find if there is a lower bound for \( a \). Since both denominators must be positive, we also check if \( 10 - 2a \) and \( 4 - 2a \) can be zero or negative.

1. \( 10 - 2a = 0 \) gives \( a = 5 \) (not included since we need \( < 5 \)).
2. \( 4 - 2a = 0 \) gives \( a = 2 \) (not included since we need \( < 2 \)).

### Step 6: Conclusion

Thus, the values of \( a \) must satisfy:

\[ -\infty < a < 2 \]

### Final Answer

The interval for \( a \) is:

\[ a \in (-\infty, 2) \]