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

Options

A. (a) a circle
B. (b) ellipse with foci on x-axis
C. (c) hyperbola with foci on x-axis
D. (d) hyperbola with foci on y-axis

Answer

To solve the problem step by step, we need to analyze the inequality given and determine the nature of the conic section represented by the equation \( \frac{x^2}{a} + \frac{y^2}{b} = 1 \).

### Step 1: Analyze the Inequality
We start with the inequality:
\[ 2^a + 2^{4-a} < 17 \]

### Step 2: Rewrite the Inequality
We can rewrite \( 2^{4-a} \) as \( \frac{16}{2^a} \):
\[ 2^a + \frac{16}{2^a} < 17 \]

### Step 3: Substitute \( t = 2^a \)
Let \( t = 2^a \). Then, the inequality becomes:
\[ t + \frac{16}{t} < 17 \]

### Step 4: Multiply by \( t \) (assuming \( t > 0 \))
Multiplying both sides by \( t \) (which is valid since \( t > 0 \)):
\[ t^2 + 16 < 17t \]

### Step 5: Rearrange the Inequality
Rearranging gives:
\[ t^2 - 17t + 16 < 0 \]

### Step 6: Factor the Quadratic
Now we factor the quadratic:
\[ (t - 1)(t - 16) < 0 \]

### Step 7: Determine the Intervals
To find the intervals where this inequality holds, we analyze the sign of the product:
- The roots are \( t = 1 \) and \( t = 16 \).
- The product \( (t - 1)(t - 16) < 0 \) is negative between the roots, i.e., for \( 1 < t < 16 \).

### Step 8: Substitute Back for \( a \)
Since \( t = 2^a \), we have:
\[ 1 < 2^a < 16 \]
Taking logarithm base 2:
\[ 0 < a < 4 \]

### Step 9: Determine the Nature of the Conic Section
We need to analyze the equation:
\[ \frac{x^2}{a} + \frac{y^2}{b} = 1 \]
where \( a > 0 \) and \( b \) is to be determined.

### Step 10: Rewrite the Equation
We can rewrite this equation as:
\[ b \cdot x^2 + a \cdot y^2 = ab \]
This is the standard form of a conic section.

### Step 11: Determine the Condition for Hyperbola
To determine whether this is an ellipse, parabola, or hyperbola, we use the condition:
\[ h^2 - ab \]
where \( h = 0 \) (since there is no linear term). Thus:
\[ 0 - ab < 0 \implies ab > 0 \]
Since \( a > 0 \) (from the interval \( 0 < a < 4 \)), for \( ab > 0 \), \( b \) must be negative.

### Conclusion
Since \( b < 0 \), the equation \( \frac{x^2}{a} - \frac{y^2}{|b|} = 1 \) represents a hyperbola.

### Final Answer
Thus, the equation \( \frac{x^2}{a} + \frac{y^2}{b} = 1 \) represents a hyperbola.