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
- A32 9
- B18 5
- C27 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