If `(9a, 6a)` is a point bounded in the region formed by parabola `y^(2)=16x and x=9`, then

Options

A. `a in (0, 1)`
B. `alt(1)/(4)`
C. `a lt 1`
D. `0lt alt4`

Answer

To determine the values of \( a \) for which the point \( (9a, 6a) \) is bounded in the region formed by the parabola \( y^2 = 16x \) and the line \( x = 9 \), we will follow these steps:

### Step 1: Understand the Boundaries
The parabola \( y^2 = 16x \) opens to the right and intersects the line \( x = 9 \). The region we are interested in is bounded by the parabola and the vertical line \( x = 9 \).

### Step 2: Determine the Conditions for \( x \)
For the point \( (9a, 6a) \) to be within the region, the \( x \)-coordinate must satisfy:
\[ 0 < 9a < 9 \]
This simplifies to:
\[ 0 < a < 1 \]

### Step 3: Determine the Conditions for \( y \)
Next, we need to ensure that the point lies within the parabola. The point \( (9a, 6a) \) must satisfy the inequality derived from the parabola:
\[ y^2 < 16x \]
Substituting \( y = 6a \) and \( x = 9a \):
\[ (6a)^2 < 16(9a) \]
This simplifies to:
\[ 36a^2 < 144a \]
Rearranging gives:
\[ 36a^2 - 144a < 0 \]
Factoring out \( 36a \):
\[ 36a(a - 4) < 0 \]

### Step 4: Analyze the Inequality
The inequality \( 36a(a - 4) < 0 \) implies that the product of \( a \) and \( (a - 4) \) must be negative. This occurs when:
- \( a < 0 \) and \( a - 4 > 0 \) (not possible since \( a \) cannot be negative)
- \( 0 < a < 4 \)

### Step 5: Combine Conditions
From Step 2, we have \( 0 < a < 1 \), and from Step 4, we have \( 0 < a < 4 \). The more restrictive condition is:
\[ 0 < a < 1 \]

### Conclusion
Thus, the values of \( a \) for which the point \( (9a, 6a) \) is bounded in the specified region are:
\[ a \in (0, 1) \]