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

Options

A. `alt4`
B. `agt4`
C. `4ltalt10`
D. `agt10`

Answer

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 1: Identify the standard form of an ellipse
The standard form of the equation of an ellipse is

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

where \(a^2\) and \(b^2\) are positive constants. For our equation to represent an ellipse, both denominators \(10 - a\) and \(4 - a\) must be positive.

### Step 2: Set up inequalities for the denominators
To ensure that the denominators are positive, we set up the following inequalities:

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

### Step 3: Solve the inequalities

**For the first inequality:**
\[ 10 - a > 0 \implies a < 10 \]

**For the second inequality:**
\[ 4 - a > 0 \implies a < 4 \]

### Step 4: Combine the results
From the inequalities, we have:

- From \(10 - a > 0\), we get \(a < 10\).
- From \(4 - a > 0\), we get \(a < 4\).

The more restrictive condition is \(a < 4\). Therefore, for the equation to represent an ellipse, \(a\) must be less than 4.

### Conclusion
The equation \(\frac{x^2}{10-a} + \frac{y^2}{4-a} = 1\) represents an ellipse if:

\[ a < 4 \]