Find the condition that line `lx + my - n = 0 ` will be a normal to the hyperbola `x^(2)/a^(2) - y^(2)/b^(2) = 1` .

Answer

To find the condition that the line \( lx + my - n = 0 \) will be a normal to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Write the equation of the hyperbola
The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Determine the equation of the normal to the hyperbola
The equation of the normal to the hyperbola at a point \((x_0, y_0)\) can be expressed in terms of the slope \(M\):
\[ y = Mx \pm \frac{(a^2 + b^2)}{\sqrt{a^2 - b^2 M^2}} \]
where \(M\) is the slope of the normal line.

### Step 3: Rewrite the given line equation
The given line \( lx + my - n = 0 \) can be rearranged to find the slope:
\[ my = -lx + n \implies y = -\frac{l}{m}x + \frac{n}{m} \]
Thus, the slope of the line is:
\[ m' = -\frac{l}{m} \]

### Step 4: Set the slopes equal
For the line to be normal to the hyperbola, the slope \(M\) of the normal must equal the slope of the line:
\[ M = -\frac{l}{m} \]

### Step 5: Substitute \(M\) into the normal equation
Now, we substitute \(M\) into the normal equation:
\[ \frac{n}{m} = \pm \left(-\frac{l}{m}\right) \frac{(a^2 + b^2)}{\sqrt{a^2 - b^2 \left(-\frac{l}{m}\right)^2}} \]
This simplifies to:
\[ \frac{n}{m} = \pm \left(-\frac{l}{m}\right) \frac{(a^2 + b^2)}{\sqrt{a^2 - \frac{b^2 l^2}{m^2}}} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ n = \pm \left(-l\right) \frac{(a^2 + b^2)}{\sqrt{a^2 - \frac{b^2 l^2}{m^2}}} \]
Squaring both sides to eliminate the square root leads to:
\[ n^2 = l^2 \frac{(a^2 + b^2)^2}{a^2 - \frac{b^2 l^2}{m^2}} \]

### Step 7: Rearranging for the condition
Rearranging the equation yields:
\[ n^2 \left(a^2 - \frac{b^2 l^2}{m^2}\right) = l^2 (a^2 + b^2)^2 \]
This gives us the condition that must be satisfied for the line to be normal to the hyperbola.

### Final Condition
The final condition can be expressed as:
\[ n^2 \left(a^2 m^2 - b^2 l^2\right) = l^2 (a^2 + b^2)^2 \]