Find the eccentricity of the conic represente | vedclass

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(A) Given equation: $x^{2} - y^{2} - 4x + 4y + 16 = 0$Rearranging terms: $(x^{2} - 4x) - (y^{2} - 4y) = -16$Completing the square: $(x^{2} - 4x + 4) -

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$\sqrt{2}$

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(A) Given equation: $x^{2} - y^{2} - 4x + 4y + 16 = 0$Rearranging terms: $(x^{2} - 4x) - (y^{2} - 4y) = -16$Completing the square: $(x^{2} - 4x + 4) - (y^{2} - 4y + 4) = -16 + 4 - 4$$(x - 2)^{2} - (y - 2)^{2} = -16$Dividing by $-16$: $\frac{(y - 2)^{2}}{16} - \frac{(x - 2)^{2}}{16} = 1$This is the equation of a rectangular hyperbola of the form $\frac{Y^{2}}{a^{2}} - \frac{X^{2}}{b^{2}} = 1$,where $a^{2} = 16$ and $b^{2} = 16$.The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$.For a rectangular hyperbola,$a = b$,so $e = \sqrt{1 + 1} = \sqrt{2}$.

Find the eccentricity of the conic represented by $x^{2} - y^{2} - 4x + 4y + 16 = 0$.

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