In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at , then the length of its latus rectum is
- A.5
- B.10
- C.8
- D.6
Show answer and solution
Correct option: A
Solution
Since a focus is at (0, 5√3), the major axis is along the y-axis, so be = 5√3.
Given: 2b - 2a = 10, so b - a = 5.
Also b² = a² + (5√3)² = a² + 75.
Substituting b = a + 5: (a+5)² = a² + 75 → a² + 10a + 25 = a² + 75 → 10a = 50 → a = 5, b = 10.
Latus rectum = 2a²/b = 2(25)/10 = 5.
Given: 2b - 2a = 10, so b - a = 5.
Also b² = a² + (5√3)² = a² + 75.
Substituting b = a + 5: (a+5)² = a² + 75 → a² + 10a + 25 = a² + 75 → 10a = 50 → a = 5, b = 10.
Latus rectum = 2a²/b = 2(25)/10 = 5.
[JEE Main 2019 · 8 April · Shift I]