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 solve the problem, we need to find the value of \( m \) such that the line \( y = mx + 7\sqrt{3} \) is normal to the hyperbola given by

\[ \frac{x^2}{24} - \frac{y^2}{18} = 1. \]

### Step 1: Identify parameters of the hyperbola
The hyperbola is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). From the equation, we can identify:

- \( a^2 = 24 \)
- \( b^2 = 18 \)

### Step 2: Write the equation of the normal to the hyperbola
The equation of the normal to the hyperbola at any point can be given by:

\[ y = mx \pm m \frac{(a^2 + b^2)}{\sqrt{a^2 - b^2}}. \]

### Step 3: Calculate \( a^2 + b^2 \) and \( a^2 - b^2 \)
Now, we calculate:

- \( a^2 + b^2 = 24 + 18 = 42 \)
- \( a^2 - b^2 = 24 - 18 = 6 \)

### Step 4: Substitute values into the normal equation
Substituting these values into the normal equation gives:

\[ y = mx \pm m \frac{42}{\sqrt{6}}. \]

### Step 5: Simplify the normal equation
Calculating \( \frac{42}{\sqrt{6}} \):

\[ \frac{42}{\sqrt{6}} = \frac{42 \sqrt{6}}{6} = 7\sqrt{6}. \]

Thus, the normal equation can be expressed as:

\[ y = mx \pm 7m\sqrt{6}. \]

### Step 6: Set the constant terms equal
We know the line \( y = mx + 7\sqrt{3} \) is normal to the hyperbola. Therefore, we can set the constant terms equal:

\[ 7m\sqrt{6} = 7\sqrt{3}. \]

### Step 7: Solve for \( m \)
Dividing both sides by 7 gives:

\[ m\sqrt{6} = \sqrt{3}. \]

Now, squaring both sides:

\[ m^2 \cdot 6 = 3. \]

This simplifies to:

\[ m^2 = \frac{3}{6} = \frac{1}{2}. \]

### Step 8: Find \( m \)
Taking the square root gives:

\[ m = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}. \]

### Final Answer
Thus, the value of \( m \) is:

\[ m = \frac{1}{\sqrt{2}} \text{ or } m = -\frac{1}{\sqrt{2}}. \]