Hyperbola Geometry in 59 Seconds — Smart Focus-Based Method ⚡
❓ Question Consider the hyperbola x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 having one of its foci at P ( − 3 , 0 ) . P(-3,\,0). If the latus rectum through its other focus subtends a right angle at point P P P , and a 2 b 2 = α 2 − β , α , β ∈ N , a^2b^2=\alpha\sqrt{2}-\beta, \quad \alpha,\beta\in\mathbb{N}, then the value of α + β \alpha+\beta is equal to ? 🖼️ Question Image ✍️ Short Solution This problem is a classic JEE Advanced–type geometry–algebra mix involving: ✔ Properties of a hyperbola ✔ Geometry of the latus rectum ✔ Right-angle condition using vectors or slopes We proceed step by step. 🔹 Step 1 — Identify hyperbola parameters Given: x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 For this hyperbola: Centre = ( 0 , 0 ) (0,0) Foci = ( ± c , 0 ) (\pm c,0) Where: c 2 = a 2 + b 2 c^2=a^2+b^2 One focus is at P ( − 3 , 0 ) P(-3,0) , so: c = 3 ⇒ a 2 + b 2 = 9 ( 1 ) c=3 \Rightarrow a^2+b^2=9 \quad (1) 🔹 Step 2 — ...