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In a rectangular hyperbola x^(2)-y^(2)=a...

In a rectangular hyperbola `x^(2)-y^(2)=a^(2)` , prove that `SP*S'P=CP^(2)` where P is any point on the hyperbola, C is origin and S and S' are foci.

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To prove that \( SP \cdot S'P = CP^2 \) for the rectangular hyperbola given by the equation \( x^2 - y^2 = a^2 \), where \( P \) is any point on the hyperbola, \( C \) is the origin, and \( S \) and \( S' \) are the foci, we can follow these steps: ### Step 1: Identify the Foci For the rectangular hyperbola \( x^2 - y^2 = a^2 \), the foci \( S \) and \( S' \) are located at \( (ae, 0) \) and \( (-ae, 0) \) respectively, where \( e \) is the eccentricity. For a rectangular hyperbola, \( e = \sqrt{2} \), so the foci are: \[ S = (a\sqrt{2}, 0), \quad S' = (-a\sqrt{2}, 0) \] ...
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