If the axes of coordinates are rotated through an angle 45° in the xy-plane keeping origin fixed, the equation $$x^2-y^2=a^2$$ changes to where (x', y') are new coordinates of (x, y).

Correct option is C

Solution:

​$${\begin{aligned}&1.\ \textbf{Rotation of Axes Formulas:} \\&\text{For rotation by angle } \theta: \\&\quad x = x'\cos\theta - y'\sin\theta \\&\quad y = x'\sin\theta + y'\cos\theta \\[5pt]&\text{For } \theta = 45^\circ: \\&\quad \cos 45^\circ = \sin 45^\circ = \frac{1}{\sqrt{2}} \\&\quad \Rightarrow x = \frac{x' - y'}{\sqrt{2}}, \quad y = \frac{x' + y'}{\sqrt{2}} \\[10pt]&2.\ \textbf{Substitute into Original Equation:} \\&\quad x^2 - y^2 = \left(\frac{x' - y'}{\sqrt{2}}\right)^2 - \left(\frac{x' + y'}{\sqrt{2}}\right)^2 = a^2 \\[5pt]&3.\ \textbf{Expand and Simplify:} \\&\quad \frac{(x' - y')^2}{2} - \frac{(x' + y')^2}{2} = a^2 \\&\quad \Rightarrow \frac{x'^2 - 2x'y' + y'^2 - (x'^2 + 2x'y' + y'^2)}{2} = a^2 \\&\quad \Rightarrow \frac{-4x'y'}{2} = a^2 \\&\quad \Rightarrow -2x'y' = a^2 \\&\quad \Rightarrow 2x'y' + a^2 = 0 \\[10pt]&\textbf{Conclusion:} \\&\text{The transformed equation is } 2x'y' + a^2 = 0, \text{ which corresponds to option C.} \\[10pt]&\textbf{Final Answer:} \\&\boxed{C}\end{aligned}}$$

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