The point (2a, a) lies inside the region bounded by the parabola `x^(2) = 4y` and its latus rectum. Then,

Options

A. `0 lt= a lt=1`
B. `0 lt a lt 1`
C. `a gt 1`
D. `a lt 0 `

Answer

To solve the problem, we need to determine the conditions under which the point (2a, a) lies inside the region bounded by the parabola \(x^2 = 4y\) and its latus rectum.

### Step-by-Step Solution:

1. **Identify the Parabola and Latus Rectum**:
The given parabola is \(x^2 = 4y\). The vertex of this parabola is at the origin (0,0), and it opens upwards. The latus rectum of this parabola is a horizontal line that passes through the focus of the parabola. The focus of the parabola \(x^2 = 4y\) is at the point (0, 1). Therefore, the equation of the latus rectum is \(y = 1\).

2. **Condition for the Point to be Inside the Parabola**:
For the point (2a, a) to lie inside the parabola, it must satisfy the inequality derived from the parabola's equation:
\[ x^2 < 4y \]
Substituting \(x = 2a\) and \(y = a\):
\[ (2a)^2 < 4a \]
Simplifying this gives:
\[ 4a^2 < 4a \]
Dividing both sides by 4 (assuming \(a \neq 0\)):
\[ a^2 < a \]
Rearranging this, we get:
\[ a^2 - a < 0 \]
Factoring out \(a\):
\[ a(a - 1) < 0 \]

3. **Finding the Intervals for \(a\)**:
The inequality \(a(a - 1) < 0\) holds true when \(a\) is between the roots of the equation \(a = 0\) and \(a = 1\). Therefore:
\[ 0 < a < 1 \]

4. **Condition for the Point to be Below the Latus Rectum**:
The point (2a, a) must also lie below the latus rectum \(y = 1\). This gives us the condition:
\[ a < 1 \]
This condition is already satisfied by the previous result.

5. **Combining Conditions**:
From the two conditions derived:
- \(0 < a < 1\)
- \(a < 1\)

We conclude that:
\[ 0 < a < 1 \]

6. **Final Conclusion**:
The point (2a, a) lies inside the region bounded by the parabola \(x^2 = 4y\) and its latus rectum if \(a\) is in the interval \( (0, 1) \).