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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

The equation of the directrices of the hyperbola 3 x^ 2 -3 y^ 2 -18...

The equation of the directrices of the hyperbola 3x2−3y2−18x+12y+2=03 x^{2}-3 y^{2}-18 x+12 y+2=0 is

Options

  1. Ax=3±136x=3 \pm \sqrt{\frac{13}{6}}
  2. Bx=3±613x=3 \pm \sqrt{\frac{6}{13}}
  3. Cx=6±133x=6 \pm \sqrt{\frac{13}{3}}
  4. Dx=6±313x=6 \pm \sqrt{\frac{3}{13}}

Correct answer

A. x=3±136x=3 \pm \sqrt{\frac{13}{6}}

Step-by-step solution

Given equation is 3 x^{2}-3 y^{2}-18 x+12 y+2=0 It can be written as (x−3)2(133)2−(y−2)2(133)2=1\frac{(x-3)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}-\frac{(y-2)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}=1 here, a=b=133∴e=1+b2a2=1+1=2\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 ±136\pm \sqrt{\frac{13}{6}}
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