Eliminate θ between the given equations:

$\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1, \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1$

$\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta = 1…..(1)\\[1em] \Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta = -1…..(2)$

Square and add equations (1) and (2):

$\Rightarrow \Big[\Big(\dfrac{x}{a}\Big) \cos \theta + \Big(\dfrac{y}{b}\Big) \sin \theta\Big]^2 + \Big[\Big(\dfrac{x}{a}\Big) \sin \theta - \Big(\dfrac{y}{b}\Big) \cos \theta\Big]^2 = 1^2 + (-1)^2 \\[1em] \Rightarrow \Big[\Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + 2\dfrac{xy}{ab} \sin \theta \cos \theta \Big] + \Big[\Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta - 2\dfrac{xy}{ab} \sin \theta \cos \theta\Big] = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 \cos^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \sin^2 \theta + \Big(\dfrac{x}{a}\Big)^2 \sin^2 \theta + \Big(\dfrac{y}{b}\Big)^2 \cos^2 \theta = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 (\cos^2 \theta + \sin^2 \theta) + \Big(\dfrac{y}{b}\Big)^2 (\sin^2 \theta + \cos^2 \theta) = 2 \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2 .$

Hence, the required relation is $\Big(\dfrac{x}{a}\Big)^2 + \Big(\dfrac{y}{b}\Big)^2 = 2$.