If the line ` y = mx + 7 sqrt(3)` is normal to the hyperbola `(x^(2))/(24)-(y^(2))/(18)=1`, then a value of m is :

Options

A. `(2)/(sqrt(5))`
B. `(sqrt(5))/(2)`
C. `(sqrt(15))/(2)`
D. `(3)/(sqrt(5))`

Explanation

To find the value of \( m \) such that the line \( y = mx + 7\sqrt{3} \) is normal to the hyperbola \( \frac{x^2}{24} - \frac{y^2}{18} = 1 \), we can follow these steps:

### Step 1: Identify parameters of the hyperbola
The given hyperbola is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Here, we have:
- \( a^2 = 24 \) → \( a = \sqrt{24} = 2\sqrt{6} \)
- \( b^2 = 18 \) → \( b = \sqrt{18} = 3\sqrt{2} \)

### Step 2: Write the normal equation for the hyperbola
The equation of the normal to the hyperbola at a point \( (x_0, y_0) \) is given by:
\[ y - y_0 = -\frac{b^2}{a^2} \cdot \frac{x - x_0}{y_0} \]
However, we can also use the formula for the normal line:
\[ y = mx + \left( \pm m \frac{a^2 + b^2}{\sqrt{a^2 - b^2}} \right) \]
Here, \( c = \pm m \frac{a^2 + b^2}{\sqrt{a^2 - b^2}} \).

### Step 3: Calculate \( a^2 + b^2 \) and \( a^2 - b^2 \)
Calculate \( a^2 + b^2 \):
\[ a^2 + b^2 = 24 + 18 = 42 \]
Calculate \( a^2 - b^2 \):
\[ a^2 - b^2 = 24 - 18 = 6 \]

### Step 4: Substitute into the normal equation
The normal line can be written as:
\[ y = mx + \left( \pm m \frac{42}{\sqrt{6}} \right) \]
We know that the line is given as \( y = mx + 7\sqrt{3} \). Therefore, we can equate:
\[ \pm m \frac{42}{\sqrt{6}} = 7\sqrt{3} \]

### Step 5: Solve for \( m \)
Taking the positive case:
\[ m \frac{42}{\sqrt{6}} = 7\sqrt{3} \]
Rearranging gives:
\[ m = \frac{7\sqrt{3} \cdot \sqrt{6}}{42} \]
Simplifying:
\[ m = \frac{7 \cdot \sqrt{18}}{42} = \frac{7 \cdot 3\sqrt{2}}{42} = \frac{21\sqrt{2}}{42} = \frac{\sqrt{2}}{2} \]

Taking the negative case:
\[ -m \frac{42}{\sqrt{6}} = 7\sqrt{3} \]
This leads to:
\[ m = -\frac{7\sqrt{3} \cdot \sqrt{6}}{42} = -\frac{\sqrt{2}}{2} \]

### Final Answer
Thus, the possible values of \( m \) are:
\[ m = \frac{\sqrt{2}}{2} \quad \text{or} \quad m = -\frac{\sqrt{2}}{2} \]