Tough JEE Question? Convert Parametric tan–Form to Line Equation FAST ⚡
❓ Question FOR: If for θ ∈ [ − π 3 , 0 ] , ( x , y ) = ( 3 tan ( θ + π 3 ) , 2 tan ( θ + π 6 ) ) lie on the curve x y + α x + β y + γ = 0 , xy + \alpha x + \beta y + \gamma = 0, then the value of α 2 + β 2 + γ 2 \alpha^2 + \beta^2 + \gamma^2 is equal to ? 🖼️ Question Image ✍️ Short Solution We are told that for every θ \theta in the interval, x = 3 tan ( θ + π 3 ) , y = 2 tan ( θ + π 6 ) x = 3\tan\left(\theta + \frac{\pi}{3}\right), \quad y = 2\tan\left(\theta + \frac{\pi}{6}\right) always satisfies x y + α x + β y + γ = 0. xy + \alpha x + \beta y + \gamma = 0. That means there is a fixed relation between x x x and y y y which does not involve θ \theta . We’ll eliminate θ \theta using a standard tangent identity. 🔹 Step 1 — Define new angles and variables Let A = θ + π 3 , B = θ + π 6 . A = \theta + \frac{\pi}{3}, \quad B = \theta + \frac{\pi}{6}. Then, x = 3 tan A , y = 2 tan B . x = 3\tan A,\quad y = 2\tan B. Compu...