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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

The equation of the hyperbola having eccentricity 2 ,~such that the...

The equation of the hyperbola having eccentricity 22,~such that the distance between its foci is 88~and the major axis is along the xx-axis, while the minor axis is along theyy-axis, is&

Options

  1. Ax212−y24=1\dfrac{x2}{12} - \dfrac{y2}{4} = 1
  2. Bx24−y212=1\dfrac{x2}{4} - \dfrac{y2}{12} = 1
  3. Cx28−y22=1\dfrac{x2}{8} - \dfrac{y2}{2} = 1
  4. Dx216−y29=1\dfrac{x2}{16} - \dfrac{y2}{9} = 1

Correct answer

B. x24−y212=1\dfrac{x2}{4} - \dfrac{y2}{12} = 1

Step-by-step solution

Distance between foci& =8=2ae= 8 = 2ae

But~e=2e = 2. Therefore,~2a=42a = 4

⇒a=2⇒a2=4⇒ a = 2 ⇒ a^{2} = 4

And~b2=a2(e2−1)=4(4−1)=12b^{2} = a^{2}\left(e^{2} - 1\right) = 4\left(4 - 1\right) = 12

∴∴Equation of hyperbola is~x24−y212=1\dfrac{x2}{4} - \dfrac{y2}{12} = 1

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