The line `lx + my+n=0` will be a normal to the hyperbola `b^2x^2-a^2y^2=a^2b^2` if

Options

A. `(a^2)/(l^2)-(b^2)/(m^2)=((a^2+b^2)^2)/(n^2)`
B. `(l^2)/(a^2)-(m^2)/(b^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. `(l^2)/(a^2)+(m^2)/(b^2)=((a^2-b^2)^2)/(n^2)`

Answer

To determine the condition under which the line \( lx + my + n = 0 \) is a normal to the hyperbola \( b^2x^2 - a^2y^2 = a^2b^2 \), we will follow these steps:

### Step 1: Rewrite the Hyperbola in Standard Form
The given hyperbola is \( b^2x^2 - a^2y^2 = a^2b^2 \). We can rewrite it in standard form by dividing through by \( a^2b^2 \):

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Write the Equation of the Normal
The equation of the normal to the hyperbola at the point \( (a \sin \phi, b \tan \phi) \) is given by:

\[ a x \sin \phi + b y = a^2 + b^2 \tan \phi \]

### Step 3: Set Up the Comparison
We need to compare this normal equation with the line equation \( lx + my + n = 0 \). Rearranging the line equation gives:

\[ lx + my = -n \]

### Step 4: Compare Coefficients
From the normal equation, we can identify the coefficients:

- Coefficient of \( x \): \( a \sin \phi \)
- Coefficient of \( y \): \( b \)
- Constant term: \( a^2 + b^2 \tan \phi \)

Setting these equal to the coefficients from the line equation, we have:

1. \( a \sin \phi = l \)
2. \( b = m \)
3. \( a^2 + b^2 \tan \phi = -n \)

### Step 5: Solve for \( \sin \phi \) and \( \tan \phi \)
From the first equation, we can express \( \sin \phi \):

\[ \sin \phi = \frac{l}{a} \]

From the second equation, we have:

\[ m = b \]

From the third equation, substituting \( \tan \phi = \frac{\sin \phi}{\cos \phi} \):

\[ \tan \phi = \frac{l}{a \cos \phi} \]

Substituting into the third equation gives:

\[ a^2 + b^2 \left( \frac{l}{a \cos \phi} \right) = -n \]

### Step 6: Express \( \cos \phi \)
Rearranging gives:

\[ \frac{b^2 l}{a \cos \phi} = -n - a^2 \]

Thus, we can express \( \cos \phi \):

\[ \cos \phi = \frac{b^2 l}{a(-n - a^2)} \]

### Step 7: Use the Identity \( \sin^2 \phi + \cos^2 \phi = 1 \)
Substituting \( \sin \phi \) and \( \cos \phi \) into the identity:

\[ \left( \frac{l}{a} \right)^2 + \left( \frac{b^2 l}{a(-n - a^2)} \right)^2 = 1 \]

### Step 8: Simplify the Equation
This leads to:

\[ \frac{l^2}{a^2} + \frac{b^4 l^2}{a^2(n + a^2)^2} = 1 \]

Multiplying through by \( a^2(n + a^2)^2 \):

\[ l^2(n + a^2)^2 + b^4 l^2 = a^2(n + a^2)^2 \]

### Step 9: Rearranging the Terms
Rearranging gives:

\[ l^2 \left( (n + a^2)^2 + b^4 \right) = a^2(n + a^2)^2 \]

### Final Condition
Thus, we arrive at the condition for the line to be a normal to the hyperbola:

\[ \frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2} \]