For what value of `lamda,` the line y=2x+`lamda` touches the hyperbola `9x^(2)-5y^(2)=45` ?

Options

A. `sqrt(11)`
B. 11
C. `-11`
D. `-sqrt(11)`

Answer

To find the value of \(\lambda\) such that the line \(y = 2x + \lambda\) touches the hyperbola \(9x^2 - 5y^2 = 45\), we can follow these steps:

### Step 1: Convert the hyperbola to standard form
The given hyperbola is:
\[ 9x^2 - 5y^2 = 45 \]
Dividing both sides by 45, we get:
\[ \frac{x^2}{5} - \frac{y^2}{9} = 1 \]
This is the standard form of the hyperbola.

### Step 2: Identify the parameters of the hyperbola
From the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can identify:
- \(a^2 = 5\) (thus \(a = \sqrt{5}\))
- \(b^2 = 9\) (thus \(b = 3\))

### Step 3: Write the equation of the tangent line
The equation of the tangent line to the hyperbola in terms of slope \(m\) is given by:
\[ y = mx \pm \sqrt{b^2m^2 - a^2} \]
Substituting \(a^2\) and \(b^2\):
\[ y = mx \pm \sqrt{9m^2 - 5} \]

### Step 4: Set the slope equal to 2
Since the line given is \(y = 2x + \lambda\), we have the slope \(m = 2\). Substituting \(m = 2\) into the tangent equation:
\[ y = 2x \pm \sqrt{9(2^2) - 5} \]
Calculating the expression under the square root:
\[ 9(4) - 5 = 36 - 5 = 31 \]
Thus, the tangent line becomes:
\[ y = 2x \pm \sqrt{31} \]

### Step 5: Relate the tangent line to the given line
The line \(y = 2x + \lambda\) touches the hyperbola if it can be expressed in the form of the tangent line. Therefore, we equate:
\[ \lambda = \pm \sqrt{31} \]

### Step 6: Final values of \(\lambda\)
Thus, the values of \(\lambda\) are:
\[ \lambda = \sqrt{31} \quad \text{or} \quad \lambda = -\sqrt{31} \]

### Conclusion
The values of \(\lambda\) for which the line touches the hyperbola are:
\[ \lambda = \pm \sqrt{31} \]

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