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

Options

A. `klt4`
B. `kgt10`
C. `4ltklt10`
D. `5in(4,7)mu(7,10)`

Answer

To determine the conditions under which the equation

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

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

### Step 1: Identify the conditions for an ellipse
The general form of the equation of an ellipse is

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

where \(a^2 > 0\) and \(b^2 > 0\). This means both denominators must be positive.

### Step 2: Set up inequalities for the denominators
From the given equation, we have:
- The first denominator: \(10 - k\)
- The second denominator: \(k - 4\)

We need both of these to be greater than zero:

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

### Step 3: Solve the inequalities

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

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

### Step 4: Combine the inequalities
From the two inequalities, we can combine them to find the range of \(k\):
\[ 4 < k < 10 \]

### Conclusion
Thus, the equation represents an ellipse if \(k\) lies in the interval \( (4, 10) \).

### Final Answer
The equation represents an ellipse if \( k \) is in the range \( 4 < k < 10 \).

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