If two distinct tangents can be drawn from the Point `(alpha,2)` on different branches of the hyperbola `x^2/9-y^2/(16)=1` then (1) `|alpha| lt 3/2` (2) `|alpha| gt 2/3` (3)`|alpha| gt 3` (4) `alpha =1`

Options

A. `|alpha|lt3//2`
B. `|alpha|gt2//3`
C. `|alpha|gt3`
D. none of these

Explanation

To solve the problem, we need to find the range of \( \alpha \) such that two distinct tangents can be drawn from the point \( (\alpha, 2) \) to the hyperbola given by the equation:

\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]

### Step 1: Identify the Hyperbola Parameters
The hyperbola can be rewritten in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where:
- \( a^2 = 9 \) so \( a = 3 \)
- \( b^2 = 16 \) so \( b = 4 \)

### Step 2: Determine the Asymptotes
The equations of the asymptotes for the hyperbola are given by:

\[ y = \pm \frac{b}{a} x \]

Substituting the values of \( a \) and \( b \):

\[ y = \pm \frac{4}{3} x \]

### Step 3: Find the Conditions for the Point \( (\alpha, 2) \)
For the point \( (\alpha, 2) \) to lie between the asymptotes, it must satisfy the following inequalities:

1. \( 2 < \frac{4}{3} \alpha \)
2. \( 2 > -\frac{4}{3} \alpha \)

### Step 4: Solve the First Inequality
From the first inequality:

\[ 2 < \frac{4}{3} \alpha \]

Multiplying both sides by \( \frac{3}{4} \):

\[ \frac{3}{2} < \alpha \quad \text{or} \quad \alpha > \frac{3}{2} \]

### Step 5: Solve the Second Inequality
From the second inequality:

\[ 2 > -\frac{4}{3} \alpha \]

Multiplying both sides by \( -\frac{3}{4} \) (remember to flip the inequality):

\[ -\frac{3}{2} < \alpha \quad \text{or} \quad \alpha > -\frac{3}{2} \]

### Step 6: Combine the Results
Now we have two inequalities:

1. \( \alpha > \frac{3}{2} \)
2. \( \alpha > -\frac{3}{2} \)

The more restrictive condition is \( \alpha > \frac{3}{2} \).

### Conclusion
Thus, the condition for \( \alpha \) such that two distinct tangents can be drawn from the point \( (\alpha, 2) \) to the hyperbola is:

\[ |\alpha| > \frac{3}{2} \]