The equation of the directrices of the hyperb | vedclass

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Question

(A) The given equation is $3x^{2}-3y^{2}-18x+12y+2=0$. Rearranging the terms: $3(x^{2}-6x) - 3(y^{2}-4y) = -2$. Completing the square: $3(x-3)^{2} - 9

Vedclass · Accepted Answer

$x=3 \pm \sqrt{\frac{13}{6}}$

Vedclass · Answer

(A) The given equation is $3x^{2}-3y^{2}-18x+12y+2=0$. Rearranging the terms: $3(x^{2}-6x) - 3(y^{2}-4y) = -2$. Completing the square: $3(x-3)^{2} - 9 - 3(y-2)^{2} + 12 = -2$. $3(x-3)^{2} - 3(y-2)^{2} = -2 + 9 - 12 = -5$. Dividing by $-5$: $\frac{(y-2)^{2}}{5/3} - \frac{(x-3)^{2}}{5/3} = 1$. This is a rectangular hyperbola with $a^{2} = 5/3$ and $b^{2} = 5/3$. The eccentricity $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1+1} = \sqrt{2}$. The directrices for a vertical hyperbola $\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1$ are $y = k \pm \frac{a}{e}$. Here $y = 2 \pm \frac{\sqrt{5/3}}{\sqrt{2}} = 2 \pm \sqrt{\frac{5}{6}}$. Note: The provided options seem to assume a horizontal orientation or a different constant. Re-evaluating the original equation: $3(x^{2}-6x+9) - 3(y^{2}-4y+4) = -2 + 27 - 12 = 13$. $3(x-3)^{2} - 3(y-2)^{2} = 13$. $\frac{(x-3)^{2}}{13/3} - \frac{(y-2)^{2}}{13/3} = 1$. Here $a^{2} = 13/3, b^{2} = 13/3, e = \sqrt{2}$. The directrices are $x = h \pm \frac{a}{e} = 3 \pm \frac{\sqrt{13/3}}{\sqrt{2}} = 3 \pm \sqrt{\frac{13}{6}}$. Thus, the correct option is $A$.

The equation of the directrices of the hyperbola $3x^{2}-3y^{2}-18x+12y+2=0$ is

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