If from the point `(alpha,alpha^2)` two tangents drawn to any one branch of hyperbola `x^2-4y^2=1`, then :

Options

A. `-(1)/(2) lt alpha lt (1)/(2)`
B. `0 lt alpha lt (1)/(2)`
C. `-1 lt alpha lt 1`
D. None of these

Answer

To solve the problem of finding the range of \( \alpha \) such that two tangents can be drawn from the point \( (\alpha, \alpha^2) \) to the hyperbola given by the equation \( x^2 - 4y^2 = 1 \), we can follow these steps:

### Step 1: Understand the Hyperbola
The equation of the hyperbola can be rewritten in standard form:
\[ \frac{x^2}{1} - \frac{y^2}{\frac{1}{4}} = 1 \]
From this, we identify \( a^2 = 1 \) and \( b^2 = \frac{1}{4} \). Thus, \( a = 1 \) and \( b = \frac{1}{2} \).

### Step 2: Find the Equation of the Tangent
The equation of the tangent to the hyperbola \( x^2 - 4y^2 = 1 \) at a point \( (x_0, y_0) \) on the hyperbola is given by:
\[ \frac{xx_0}{1} - \frac{yy_0}{\frac{1}{4}} = 1 \]
Simplifying this gives:
\[ xx_0 - 4yy_0 = 1 \]

### Step 3: Substitute the Point \( (\alpha, \alpha^2) \)
We will substitute \( x = \alpha \) and \( y = \alpha^2 \) into the tangent equation:
\[ \alpha x_0 - 4\alpha^2 y_0 = 1 \]

### Step 4: Condition for Tangents
For the tangents to exist, the discriminant of the resulting quadratic equation in \( x_0 \) or \( y_0 \) must be non-negative.

### Step 5: Use the Asymptotes
The asymptotes of the hyperbola are given by:
\[ y = \pm \frac{1}{2}x \]
This means that the tangents from the point \( (\alpha, \alpha^2) \) must lie within the bounds defined by these asymptotes.

### Step 6: Find the Intersection with Asymptotes
To find the range of \( \alpha \), we set up the equations of the lines:
1. \( y = \frac{1}{2}x \)
2. \( y = -\frac{1}{2}x \)

Substituting \( y = \alpha^2 \) into these gives:
1. \( \alpha^2 = \frac{1}{2}\alpha \)
2. \( \alpha^2 = -\frac{1}{2}\alpha \)

### Step 7: Solve for \( \alpha \)
1. From \( \alpha^2 = \frac{1}{2}\alpha \):
\[ \alpha^2 - \frac{1}{2}\alpha = 0 \implies \alpha(\alpha - \frac{1}{2}) = 0 \]
Thus, \( \alpha = 0 \) or \( \alpha = \frac{1}{2} \).

2. From \( \alpha^2 = -\frac{1}{2}\alpha \):
\[ \alpha^2 + \frac{1}{2}\alpha = 0 \implies \alpha(\alpha + \frac{1}{2}) = 0 \]
Thus, \( \alpha = 0 \) or \( \alpha = -\frac{1}{2} \).

### Step 8: Combine the Results
The values of \( \alpha \) that allow for the tangents to exist are:
\[ -\frac{1}{2} \leq \alpha \leq \frac{1}{2} \]

### Final Answer
Thus, the range of \( \alpha \) is:
\[ \alpha \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]