If `m_(1)` and `m_(2)` are two values of m for which the line y = `mx+ 2sqrt(5)` is a tangent to the hyperbola `(x^(2))/(4)-(y^(2))/(16)=1` then the value of `|m_(1)+(1)/(m_(2))|` is equal to

Options

A. `8/3`
B. `10/3`
C. 0
D. 9

Answer

To solve the problem, we need to find the values of \( m_1 \) and \( m_2 \) for which the line \( y = mx + 2\sqrt{5} \) is a tangent to the hyperbola \( \frac{x^2}{4} - \frac{y^2}{16} = 1 \). Then, we will compute \( |m_1 + \frac{1}{m_2}| \).

### Step 1: Identify the equation of the hyperbola and the line
The hyperbola is given by:
\[ \frac{x^2}{4} - \frac{y^2}{16} = 1 \]
The line is given by:
\[ y = mx + 2\sqrt{5} \]

### Step 2: Substitute the line equation into the hyperbola equation
Substituting \( y = mx + 2\sqrt{5} \) into the hyperbola equation:
\[ \frac{x^2}{4} - \frac{(mx + 2\sqrt{5})^2}{16} = 1 \]

### Step 3: Expand and simplify
Expanding the equation:
\[ \frac{x^2}{4} - \frac{m^2x^2 + 4m\sqrt{5}x + 20}{16} = 1 \]
Multiply through by 16 to eliminate the denominators:
\[ 4x^2 - (m^2x^2 + 4m\sqrt{5}x + 20) = 16 \]
This simplifies to:
\[ (4 - m^2)x^2 - 4m\sqrt{5}x - 36 = 0 \]

### Step 4: Condition for tangency
For the line to be tangent to the hyperbola, the discriminant of this quadratic equation must be zero:
\[ D = b^2 - 4ac = 0 \]
Here, \( a = 4 - m^2 \), \( b = -4m\sqrt{5} \), and \( c = -36 \).

Calculating the discriminant:
\[ (-4m\sqrt{5})^2 - 4(4 - m^2)(-36) = 0 \]
\[ 16 \cdot 5m^2 + 144(4 - m^2) = 0 \]
\[ 80m^2 + 576 - 144m^2 = 0 \]
\[ -64m^2 + 576 = 0 \]

### Step 5: Solve for \( m^2 \)
Rearranging gives:
\[ 64m^2 = 576 \]
\[ m^2 = \frac{576}{64} = 9 \]
Thus, \( m = \pm 3 \). Therefore, we have \( m_1 = 3 \) and \( m_2 = -3 \).

### Step 6: Calculate \( |m_1 + \frac{1}{m_2}| \)
Now we compute:
\[ |m_1 + \frac{1}{m_2}| = |3 + \frac{1}{-3}| \]
\[ = |3 - \frac{1}{3}| \]
\[ = |3 - 0.3333| = |2.6667| = \frac{8}{3} \]

Thus, the final answer is:
\[ \boxed{\frac{8}{3}} \]