The equation of the directrices of the conic | vedclass

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(C) Given equation: $x^2 + 2x - y^2 + 5 = 0$Completing the square for $x$: $(x^2 + 2x + 1) - y^2 + 5 - 1 = 0$$(x + 1)^2 - y^2 = -4$Dividing by $-4$: $

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$y = \pm \sqrt{2}$

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(C) Given equation: $x^2 + 2x - y^2 + 5 = 0$Completing the square for $x$: $(x^2 + 2x + 1) - y^2 + 5 - 1 = 0$$(x + 1)^2 - y^2 = -4$Dividing by $-4$: $\frac{y^2}{4} - \frac{(x + 1)^2}{4} = 1$This is a hyperbola of the form $\frac{y^2}{b^2} - \frac{(x-h)^2}{a^2} = 1$,where $a^2 = 4$ and $b^2 = 4$.Here,$a = 2$ and $b = 2$.The eccentricity $e = \sqrt{1 + \frac{a^2}{b^2}} = \sqrt{1 + \frac{4}{4}} = \sqrt{2}$.The equations of the directrices for this hyperbola are $y - k = \pm \frac{b}{e}$.Since $k = 0$,$y = \pm \frac{2}{\sqrt{2}} = \pm \sqrt{2}$.

The equation of the directrices of the conic $x^2 + 2x - y^2 + 5 = 0$ are

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