PQ is a double ordinate of the hyperbola x2a2−y2b2=1 such that OPQ is an equilateral triangle, O being the centre of the hyperbola, then the range of the eccentricity e of the hyperbola is
A
1<e<2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1<e<√2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
e≥2√3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
e<4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ce≥2√3
Let double ordinate PQ be such that P=(asecθ,btanθ) and Q=(asecθ,−btanθ) and O is the centre (0,0).
From △OPR tan30∘=btanθasecθ ⇒1√3=basinθ ⇒3b2a2=cosec2θ ⇒3(e2−1)=cosec2θ≥1 ⇒e≥2√3