Sides of an equilateral triangle ABC touch the parabola y²=4ax, then the points A,B and C lie on

1. y²=(x+a)² +4ax

1. y²=3(x+a)² + ax

1. y²=3(x+a)² +4ax

d.y²=(x+a)² +ax

2 answers · Mathematics Best AnswerAny tangent to y²=4ax is y=mx+a/m where m=tan(θ) and θ is inclination to x-axis. Any two tangents of gradients m₁ & m₂ intersect when m₁x+a/m₁ = m₂x+a/m₂ This gives for their intersection x=a/(m₁m₂) … (i) y=a(1/m₁+1/m₂) … (ii) If these two tangents are the sides of an equilateral Δ then θ₂−θ₁=±60° → { tan(θ₂)−tan(θ₁) } / { 1+tan(θ₁)tan(θ₂) } = ±tan(60°) = ±√3 → (m₂−m₁)² / (1+m₁m₂)² = 3 … (iii) From (ii) & (i) y = a(m₁+m₂)/(m₁m₂) = x(m₁+m₂) → m₁+m₂=y/x (m₂−m₁)² = (m₂+m₁)²−4m₁m₂ = y²/x² − 4a/x = (y²−4ax)/x² So (iii) becomes { (y²−4ax)/x² } / (1+a/x)² = 3 i.e. y²−4ax = 3(x+a)², a hyperbola