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

(a)  3a(2 – √3 )

(b)  4a(2 – √3 )

(c)  2(2 – √3 )

(d)  8a(2 – √3)

Answer 1

Correct option (b)  4a(2 – √3)

Explanation:

Let A = (at₁² , 2at₁), B = (at₂ ² , – 2at₁)

We have m_AS = tan (π/6) 

⇒2at₁/at₁² - a = - 1/√3

⇒ t₁² + 2√3t₁ = -1 = 0

⇒ t₁ = -√3 ± 2

Clearly t₁ = – √3 – 2 is rejected. Thus t₁ = (2 – √3). Hence AB = 4at₁ = 4a(2 - √3).