If `2^a+2^(4-a) lt 17`, then `(x^2)/(a)+(y^2)/(b)=1` reperesents

Options

A. a circle
B. ellipse with foci on x-axis
C. hyperbola with foci on x-axis
D. hyperbola with foci on y-axis

Answer

To solve the inequality \(2^a + 2^{4-a} < 17\) and determine what the equation \(\frac{x^2}{a} + \frac{y^2}{b} = 1\) represents, we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ 2^a + 2^{4-a} < 17 \]
We can rewrite \(2^{4-a}\) as \(\frac{16}{2^a}\) since \(2^{4-a} = 2^4 \cdot 2^{-a} = \frac{16}{2^a}\). Thus, we can express the inequality as:
\[ 2^a + \frac{16}{2^a} < 17 \]

### Step 2: Let \(t = 2^a\)
Let \(t = 2^a\). The inequality now becomes:
\[ t + \frac{16}{t} < 17 \]

### Step 3: Multiply through by \(t\)
To eliminate the fraction, multiply both sides by \(t\) (assuming \(t > 0\)):
\[ t^2 + 16 < 17t \]

### Step 4: Rearrange the Inequality
Rearranging gives us:
\[ t^2 - 17t + 16 < 0 \]

### Step 5: Factor the Quadratic
Next, we factor the quadratic:
\[ (t - 1)(t - 16) < 0 \]

### Step 6: Analyze the Inequality
To find the intervals where this inequality holds, we check the sign of the product:
- The roots are \(t = 1\) and \(t = 16\).
- The expression \((t - 1)(t - 16)\) is negative between the roots, so:
\[ 1 < t < 16 \]

### Step 7: Convert Back to \(a\)
Since \(t = 2^a\), we take the logarithm base 2:
\[ 1 < 2^a < 16 \]
This implies:
\[ 0 < a < 4 \]

### Step 8: Determine the Representation of the Conic Section
Now, we analyze the equation \(\frac{x^2}{a} + \frac{y^2}{b} = 1\). Given that \(a\) is positive and we need to determine \(b\):
- If \(b < 0\), the equation becomes \(\frac{x^2}{a} - \frac{y^2}{|b|} = 1\), which represents a hyperbola.

### Conclusion
Thus, the inequality \(2^a + 2^{4-a} < 17\) leads us to conclude that \(a\) is between \(0\) and \(4\), and if \(b\) is negative, the equation \(\frac{x^2}{a} + \frac{y^2}{b} = 1\) represents a hyperbola.