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QJEEM23Ma11f6: JEE Main 2023 Mathematics - Class 11 JEEM23Ma11f6 Solution: Let be the hyperbola, whose foci are and eccentricity is . Then …

Let H\mathrm{H} be the hyperbola, whose foci are (1±2,0)(1 \pm \sqrt{2}, 0) and eccentricity is 2\sqrt{2} . Then the length of its latus rectum is :

Multiple Choice Options

  • A 52\frac{5}{2}
  • B 3
  • C 2
  • D 32\frac{3}{2}

Step-by-Step Verified Solution

2ae=∣(1+2)−(1+2)∣=222 \mathrm{ae}=|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2}

⇒ ae=2\mathrm{ae}=\sqrt{2}

⇒ a=1\mathrm{a}=1

⇒b=1∵e=2⇒\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow Hyperbola is rectangular

⇒L.R=2 b2a=2\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2

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