If the point `(2a,a)` lies inside the parabola `x^(2) -2x - 4y +3 = 0`, then a lies in the interval (a) `[1/2,3/2]` (b) `(1/2,3/2)` (c) `(1,3)` (d) `(-3/2,-1/2)`

Options

A. `[(1)/(2),(3)/(2)]`
B. `((1)/(2),(3)/(2))`
C. `(1,3)`
D. `((-3)/(2),(-1)/(2))`

Explanation

To determine the interval for \( a \) such that the point \( (2a, a) \) lies inside the parabola defined by the equation \( x^2 - 2x - 4y + 3 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation of the parabola
The given equation of the parabola is:
\[ x^2 - 2x - 4y + 3 = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ 4y = x^2 - 2x + 3 \]
\[ y = \frac{1}{4}(x^2 - 2x + 3) \]

### Step 2: Substitute the point into the equation
We need to check if the point \( (2a, a) \) lies inside the parabola. To do this, we substitute \( x = 2a \) and \( y = a \) into the equation:
\[ a < \frac{1}{4}((2a)^2 - 2(2a) + 3) \]
Simplifying this:
\[ a < \frac{1}{4}(4a^2 - 4a + 3) \]
\[ a < a^2 - a + \frac{3}{4} \]

### Step 3: Rearranging the inequality
Now, we rearrange the inequality:
\[ 0 < a^2 - 2a + \frac{3}{4} \]
This can be rewritten as:
\[ a^2 - 2a + \frac{3}{4} > 0 \]

### Step 4: Finding the roots of the quadratic
To find the intervals where this quadratic is positive, we first find its roots using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = \frac{3}{4} \):
\[ a = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \frac{3}{4}}}{2 \cdot 1} \]
\[ = \frac{2 \pm \sqrt{4 - 3}}{2} \]
\[ = \frac{2 \pm 1}{2} \]
This gives us the roots:
\[ a = \frac{3}{2}, \quad a = \frac{1}{2} \]

### Step 5: Analyzing the intervals
Now we analyze the sign of the quadratic \( a^2 - 2a + \frac{3}{4} \) in the intervals determined by the roots:
1. \( (-\infty, \frac{1}{2}) \)
2. \( \left(\frac{1}{2}, \frac{3}{2}\right) \)
3. \( (\frac{3}{2}, \infty) \)

- For \( a < \frac{1}{2} \): The quadratic is positive.
- For \( \frac{1}{2} < a < \frac{3}{2} \): The quadratic is negative.
- For \( a > \frac{3}{2} \): The quadratic is positive.

### Step 6: Conclusion
Since we want the quadratic to be greater than zero, the valid intervals for \( a \) are:
\[ a \in (-\infty, \frac{1}{2}) \cup (\frac{3}{2}, \infty) \]
However, since we need \( (2a, a) \) to lie inside the parabola, we focus on the interval:
\[ \left(\frac{1}{2}, \frac{3}{2}\right) \]
Thus, the correct answer is:
\[ \text{(b) } (1/2, 3/2) \]