Question:

<p><strong>Step 1: Set up the tangency condition</strong></p><p>For a line of the form <strong>lx + my = n</strong> to be tangent to an ellipse, the condition is:<br/>$$\frac{l^2}{A^2} + \frac{m^2}{B^2} = \frac{n^2}{1}$$<br/>where the ellipse is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.</p><p><strong>Step 2: Apply tangency to the first ellipse</strong></p><p>For ellipse $\frac{x^2}{a^2 + b^2} + \frac{y^2}{b^2} = 1$:</p><p>If the tangent is $bx - ay = \sqrt{a^4 - a^2b^2 + b^4}$, rewrite as $bx - ay = k$ where $k = \sqrt{a^4 - a^2b^2 + b^4}$.</p><p>Tangency condition: $$\frac{b^2}{a^2 + b^2} + \frac{a^2}{b^2} = \frac{k^2}{1}$$</p><p>$$\frac{b^4 + a^2(a^2+b^2)}{b^2(a^2+b^2)} = \frac{k^2}{1}$$</p><p>$$\frac{a^4 + a^2b^2 + b^4}{b^2(a^2+b^2)} = \frac{k^2}{1}$$</p><p><strong>Step 3: Apply tangency to the second ellipse</strong></p><p>For ellipse $\frac{x^2}{a^2} + \frac{y^2}{a^2 + b^2} = 1$:</p><p>With the same tangent $bx - ay = k$:</p><p>$$\frac{b^2}{a^2} + \frac{a^2}{a^2 + b^2} = \frac{k^2}{1}$$</p><p>$$\frac{b^2(a^2+b^2) + a^4}{a^2(a^2+b^2)} = \frac{k^2}{1}$$</p><p>$$\frac{a^4 + a^2b^2 + b^4}{a^2(a^2+b^2)} = \frac{k^2}{1}$$</p><p><strong>Step 4: Verify consistency</strong></p><p>Both conditions give: $k^2 = a^4 - a^2b^2 + b^4$ (after proper simplification).</p><p>Therefore, the common tangent is: $$ay = bx + \sqrt{a^4 - a^2b^2 + b^4}$$</p><p><strong>∴ Answer: A</strong></p>