Step-by-step Solution:

To determine the condition for the line \( kx + my + n = 0 \) to be tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can proceed with the following steps: 1. Given Line: \[ kx + my + n = 0 \] Solve for \( y \): \[ my = -kx - n \Rightarrow y = -\frac{k}{m}x - \frac{n}{m} \] 2. Condition for Tangency: For the line to be tangent to the ellipse, the distance from the center of the ellipse to the line must satisfy: \[ c^2 = a^2 m^2 + b^2 \] where \( c \) is the distance from the center to the line. 3. Express \( c \) in Terms of \( n \): The distance \( c \) from the origin to the line \( kx + my + n = 0 \) is: \[ c = \frac{|n|}{\sqrt{k^2 + m^2}} \] Therefore: \[ c^2 = \frac{n^2}{k^2 + m^2} \] 4. Set the Tangency Condition: Equate the two expressions for \( c^2 \): \[ \frac{n^2}{k^2 + m^2} = a^2 m^2 + b^2 \] Multiply both sides by \( k^2 + m^2 \): \[ n^2 = a^2 m^2 (k^2 + m^2) + b^2 (k^2 + m^2) \] Simplify: \[ n^2 = a^2 m^2 + b^2 \] Therefore, the condition for the line \( kx + my + n = 0 \) to be tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is: \[ \boxed{n^2 = a^2 m^2 + b^2} \]