The equation `(x^(2))/(10-a)+(y^(2))/(4-a)=1` represents an ellipse , if

Options

A. `alt4`
B. `agt4`
C. `4lt a`
D. `agt10`

Explanation

To determine the conditions under which the equation

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

represents an ellipse, we need to analyze the denominators of the fractions in the equation.

### Step-by-Step Solution:

1. **Identify the General Form of an Ellipse**:
The general form of an ellipse is given by

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a^2 > 0 \) and \( b^2 > 0 \).

2. **Set Conditions for the Denominators**:
In our equation, we have the denominators \( 10 - a \) and \( 4 - a \). For the equation to represent an ellipse, both denominators must be positive:

\[ 10 - a > 0 \quad \text{and} \quad 4 - a > 0 \]

3. **Solve the Inequalities**:
- From the first inequality \( 10 - a > 0 \):
\[ a < 10 \]
- From the second inequality \( 4 - a > 0 \):
\[ a < 4 \]

4. **Combine the Conditions**:
Since both conditions must be satisfied simultaneously, we take the more restrictive condition:

\[ a < 4 \]

5. **Conclusion**:
Therefore, the equation represents an ellipse if:

\[ a < 4 \]