Question

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

• A. $x=3\pm \sqrt{\frac{13}{6}}$
• B. $x=3\pm \sqrt{\frac{6}{13}}$
• C. $x=6\pm \sqrt{\frac{13}{3}}$
• D. $x=6\pm \sqrt{\frac{3}{13}}$

Answer 1

The correct option is A $x=3\pm \sqrt{\frac{13}{6}}$

$3x^2-3y^2-18x+12y+2=0$
$\Rightarrow 3(x^2-6x)-3(y^2-4y)+2=0$
$\Rightarrow 3(x-3{)}^2-3(y-2{)}^2=-2+27-12$
$\Rightarrow (x-3{)}^2-(y-2{)}^2=\frac{13}{3}$
$\Rightarrow \frac{(x-3{)}^2}{(\sqrt{13/3}{)}^2}-\frac{(y-3{)}^2}{(\sqrt{13/3}{)}^2}=1$

Equation of directrix are $x-k=\pm \frac{a}{e}$
$\Rightarrow x-3=\pm \frac{13/3}{\sqrt{13/3+13/3}}\Rightarrow x=3\pm \sqrt{\frac{13}{6}}$