If the transformed equation of a curve is $x^{2} - 2 x y T a n 2 α - y^{2} = a^{2}$ when the axes are rotated through an angle $α$ then the original equation of the curve is

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$x^{2} - y^{2} = a^{2} \cos 2 α$

$x^{2} + y^{2} = y^{2} \cos 2 α$

$x^{2} - a^{2} = y^{2} \cos 2 α$

$x^{2} + y^{2} = a^{2} \cos 2 d$

answer is B.

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Detailed Solution

$X = x \cos α + y \sin α, Y = - x \sin α + y \cos α$
The original equation of the curve is ${(x \cos α + y \sin α)}^{2} - 2 (x \cos α + y \sin α)$
$\begin{array}{l} (- x \sin α + y \cos α \tan 2 α)^{2} - 1 (- x \sin α + y \cos α)^{2} = a^{2} \\ \Rightarrow x^{2} \cos^{2} α + y^{2} \sin^{2} α + 2 x y \cos α \sin α + 2 x^{2} \cos α \sin α \tan 2 α - 2 x y \cos^{2} α \tan 2 α \\ + 2 x y \sin^{2} α \tan 2 α - 2 y^{2} \cos α \sin α \tan 2 α - x^{2} \sin^{2} α - y^{2} \cos^{2} α + 2 x y \sin α \cos α = a^{2} \\ \Rightarrow x^{2} (\cos^{2} α + \sin 2 α \tan 2 α - \sin^{2} α) - y^{2} (- \sin^{2} α + \sin 2 α \tan 2 α + \cos^{2} α) \\ + x y (\sin 2 α - 2 \cos^{2} α \tan 2 α + 2 \sin^{2} α \tan 2 α + \sin 2 α) = a^{2} \\ \Rightarrow (x^{2} - y^{2}) (\cos 2 α + \frac{\sin 2 α}{\cos 2 α}) = a^{2} \Rightarrow x^{2} - y^{2} = a^{2} \cos 2 α \end{array}$