The line `y=mx+2` touches the hyperola `4x^(2)-9y^(2)=36` then m=

Options

A. `pm 2/3`
B. `pm (2sqrt2)/3`
C. `pm 8/9`
D. `pm (4sqrt2)/(3)`

Answer

To solve the problem, we need to find the value of \( m \) such that the line \( y = mx + 2 \) touches the hyperbola given by the equation \( 4x^2 - 9y^2 = 36 \).

### Step 1: Rewrite the hyperbola in standard form
We start with the equation of the hyperbola:
\[ 4x^2 - 9y^2 = 36 \]
Dividing both sides by 36 gives:
\[ \frac{x^2}{9} - \frac{y^2}{4} = 1 \]
This is the standard form of the hyperbola, where \( a^2 = 9 \) and \( b^2 = 4 \).

### Step 2: Identify \( a \) and \( b \)
From the standard form, we can identify:
\[ a = 3 \quad \text{and} \quad b = 2 \]

### Step 3: Use the condition for tangency
For the line \( y = mx + 2 \) to be tangent to the hyperbola, we use the condition:
\[ c^2 = a^2 m^2 - b^2 \]
where \( c \) is the y-intercept of the line. Here, \( c = 2 \).

### Step 4: Substitute values into the tangency condition
Substituting the known values into the tangency condition:
\[ 2^2 = 9m^2 - 4 \]
This simplifies to:
\[ 4 = 9m^2 - 4 \]
Adding 4 to both sides gives:
\[ 8 = 9m^2 \]

### Step 5: Solve for \( m^2 \)
Now, we solve for \( m^2 \):
\[ m^2 = \frac{8}{9} \]

### Step 6: Find \( m \)
Taking the square root of both sides gives:
\[ m = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3} \]

### Conclusion
Thus, the values of \( m \) for which the line touches the hyperbola are:
\[ m = \pm \frac{2\sqrt{2}}{3} \]

### Final Answer
The correct answer is \( m = \pm \frac{2\sqrt{2}}{3} \).

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