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. 2a (2 –$\sqrt{3}$)
• B. 4a (2 –$\sqrt{3}$)
• C. a (2 –$\sqrt{3}$)
• D. 8a (2 –$\sqrt{3}$)

Answer 1

Answer: (B)

4a (2 –$\sqrt{3}$)

Answer 2

Let A(at₁², 2at₁), B ŗ (at₁² , –2at₁). We have

m_AS = tan$\left( \frac{5\pi}{6} \right)$ Ž $\frac{2at_{1}}{at_{1}^{2} - a}$= – $\frac{1}{\sqrt{3}}$

Ž t₁² + 2$\sqrt{3}$t₁ – 1 = 0

Ž t₁ = –$\sqrt{3}$± 2.

Clearly t₁ = –$\sqrt{3}$ – 2 is rejected. Thus,

t₁ = (2 –$\sqrt{3}$)

Hence, AB = 4at₁ = 4a (2 –$\sqrt{3}$).