Q81 / 155 · WBJEE 2019 · Mathematics
The equation of the directrices of the hyperbola $3 x^{2}-3 y^{2}-18 x+12 y+2=0$ is
- Correct answerA$x=3 \pm \sqrt{\frac{13}{6}}$
- B$x=3 \pm \sqrt{\frac{6}{13}}$
- C$x=6 \pm \sqrt{\frac{13}{3}}$
- D$x=6 \pm \sqrt{\frac{3}{13}}$
Solution
Given equation is
$$
3 x^{2}-3 y^{2}-18 x+12 y+2=0
$$
It can be written as
$\frac{(x-3)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}-\frac{(y-2)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}=1$
$\begin{aligned} \text { here, } a &=b=\sqrt{\frac{13}{3}} \\ \therefore \quad e &=\sqrt{1+\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1+1} \\ &=\sqrt{2} \end{aligned}$
$\therefore$ Equation of the directrix are
$$
x=3 \pm \sqrt{\frac{13}{6}}
$$
Hence, the correct answer is option: 'A'.