For what value of `lambda` does the line `y=2x+lambda` touches the hyperbola `16x^(2)-9y^(2)=144`?

Answer

To find the value of \( \lambda \) for which the line \( y = 2x + \lambda \) touches the hyperbola \( 16x^2 - 9y^2 = 144 \), we can follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given hyperbola is:
\[ 16x^2 - 9y^2 = 144 \]
Divide both sides by 144:
\[ \frac{16x^2}{144} - \frac{9y^2}{144} = 1 \]
This simplifies to:
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]
Thus, we can identify \( a^2 = 9 \) and \( b^2 = 16 \), giving \( a = 3 \) and \( b = 4 \).

### Step 2: Identify the slope and intercept of the line
The line is given as:
\[ y = 2x + \lambda \]
From this, we can see that the slope \( m \) is \( 2 \) and the y-intercept \( c \) is \( \lambda \).

### Step 3: Use the condition for tangency
For a line to be tangent to a hyperbola, the following relationship must hold:
\[ c = \sqrt{a^2 m^2 - b^2} \]
Substituting the values we have:
- \( a = 3 \)
- \( b = 4 \)
- \( m = 2 \)

### Step 4: Substitute the values into the tangency condition
Substituting into the equation:
\[ \lambda = \sqrt{9 \cdot (2^2) - 16} \]
Calculating this gives:
\[ \lambda = \sqrt{9 \cdot 4 - 16} = \sqrt{36 - 16} = \sqrt{20} \]

### Step 5: Simplify the result
We can simplify \( \sqrt{20} \):
\[ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} \]
Thus, the value of \( \lambda \) is:
\[ \lambda = 2\sqrt{5} \]

### Final Answer
The value of \( \lambda \) for which the line touches the hyperbola is:
\[ \lambda = 2\sqrt{5} \]