The line `lx+my=n` is a normal to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` if

Options

A. `(a^(2))/(l^(2))+(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`
B. `(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)-b^(2))^(2))/(n^(2))`
C. `(a^(2))/(l^(2))-(b^(2))/(m^(2))=((a^(2)+b^(2))^(2))/(n^(2))`
D. none

Answer

To determine the condition under which the line \( lx + my = n \) is a normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Understand the Normal to the Hyperbola
The normal to the hyperbola at a point can be expressed in terms of the angle \( \phi \) as follows:
- The coordinates of the point on the hyperbola can be represented as \( (a \sec \phi, b \tan \phi) \).

### Step 2: Write the Equation of the Normal
The equation of the normal line at the point \( (a \sec \phi, b \tan \phi) \) can be derived. The normal line can be expressed as:
\[ a \sin \phi (x - a \sec \phi) - b \cos \phi (y - b \tan \phi) = 0 \]
This simplifies to:
\[ a \sin \phi \cdot x - b \cos \phi \cdot y = a^2 \sin \phi + b^2 \cos \phi \]
This can be rearranged to:
\[ a \sin \phi \cdot x + b \cos \phi \cdot y = a^2 \sin \phi + b^2 \cos \phi \]

### Step 3: Compare with the Line Equation
The line equation given is \( lx + my = n \). We can rewrite this as:
\[ lx + my - n = 0 \]
We need to ensure that the two equations represent the same line.

### Step 4: Set Up the Condition
From the two equations, we can equate coefficients:
1. \( a \sin \phi = l \)
2. \( b \cos \phi = m \)
3. \( a^2 \sin \phi + b^2 \cos \phi = n \)

### Step 5: Use Trigonometric Identity
Using the identity \( \sin^2 \phi + \cos^2 \phi = 1 \), we can express:
\[ \frac{b^2 l^2}{a^2 m^2} + \frac{(a^2 + b^2)^2}{n^2} = 1 \]

### Step 6: Rearranging the Terms
Rearranging gives us:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} \]

### Conclusion
Thus, the condition for the line \( lx + my = n \) to be a normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is:
\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} \]