Given the foci points of a hyperbola are (1±2,0)
So, S(1+2,0),S′(1−2,0)
SS′=2ae=(1+2)−(1−2)
⇒2ae=22
⇒2×a×2=22,(givene=2)
⇒a=1
Since b2=a2(e2−1)
⇒b2=1(2−1)=1
So, the latus rectum of a hyperbola is given by,
L.R. =a2b2
L.R. =2
Let H be the hyperbola, whose foci are (1±2,0) and eccentricity is 2. Then the length of its latus rectum is:
Held on 31 Jan 2023 · Verified 6 Jul 2026.
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