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JEE Main202225 Jul 2022Evening ShiftMathematicsHyperbolaActual

Let the foci of the ellipse x 2 16 + y 2 7 = 1 and the hyperbola x 2 144 - y 2 α = 1 25 coincide. Then the length of the latus rectum of the hyperbola is:

Options

  1. A32 9
  2. B18 5
  3. C27 4
  4. D27 10

Correct answer

D. 27 10

Step-by-step solution

Given equation of ellipse x 2 16 + y 2 7 = 1 Now finding eccentricity = 1 - 7 16 = 3 4 So, foci ≡ ± a e , 0 ≡ ± 3 , 0 Now, hyperbola: x 2 144 25 - y 2 α 25 = 1 Eccentricity will be = 1 + α 144 = 1 12 144 + α Foci ≡ ± a e , 0 ≡ ± 12 5 · 1 12 144 + α , 0 Given foci coincide then 3 = 1 5 144 + α ⇒ α = 81 Hence, hyperbola is x 2 12 5 2 - y 2 9 5 2 = 1 Length of latus rectum = 2 b 2 a = 2 · 81 25 12 5 = 27 10

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