Maths-
General
Easy
Question
An equilateral triangle SAB is inscribed in the parabola y2 = 4ax having it’s focus at ‘S’. If chord AB lies towards the left of S, then side length of this triangle is -
- 2a (2 –
)
- 4a (2 –)
- a (2 –)
- 8a (2 –)
)
)
)
)
The correct answer is: 4a (2 –)
Let A(at12, 2at1), B ≡ (at12 , –2at1). We have
mAS = tan
= – 
t12 + 2t1 – 1 = 0
t1 = –± 2.
Clearly t1 = – – 2 is rejected. Thus,
t1 = (2 –)
Hence, AB = 4at1 = 4a (2 –).
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