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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced · Mathematics · Hyperbola

The director circle of a hyperbola is x^2+y^2-4 y=0 . One end of...

The director circle of a hyperbola is x2+y2−4y=0x^2+y^2-4 y=0. One end of the transverse axis is (2,0)(2,0) then a focus isThe director circle of a hyperbola is x^2+y^2-4 y=0 . One end of the transverse axis is (2,0) then a focus isThe director circle of a hyperbola is x^2+y^2-4 y=0 . One end of the transverse axis is (2,0) then a focus is

Options

  1. A(3,2−3)(\sqrt{3}, 2-\sqrt{3})
  2. B(−3,2+3)(-\sqrt{3}, 2+\sqrt{3})
  3. C(6.2−6)(\sqrt{6} .2-\sqrt{6})
  4. D(−6,2+6)(-\sqrt{6}, 2+\sqrt{6})

Correct answer

D. (−6,2+6)(-\sqrt{6}, 2+\sqrt{6})

Step-by-step solution

Radius of director circle =a2−b2=2=\sqrt{\mathrm{a}^2-\mathrm{b}^2}=2(22)2−b2=4b2=4e=32\begin{aligned} & (2 \sqrt{2})^2-b^2=4 \\ & b^2=4 \quad e=\frac{\sqrt{3}}{\sqrt{2}}\end{aligned}∴\therefore Focus lies on the line x−0−12=y−212\frac{x-0}{\frac{-1}{\sqrt{2}}}=\frac{\mathrm{y}-2}{\frac{1}{\sqrt{2}}} at a distance ae units from centre. ∴x−0−12=y−212=±23(x,y)=(6.2−6)\therefore \frac{x-0}{-\frac{1}{\sqrt{2}}}=\frac{y-2}{\frac{1}{\sqrt{2}}}= \pm 2 \sqrt{3}(x, y)=(\sqrt{6} .2-\sqrt{6}) or (−6,2+6)(-\sqrt{6}, 2+\sqrt{6}) The director circle of a hyperbola is x^2+y^2-4 y=0 . One end of the transverse axis is (2,0) then a focus is
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