Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Concept 🎬 Hyperbola – Focus & Latus Rectum Concept in 59 Sec Hyperbola dikhe, focus dikhe, aur latus rectum right angle subtend kar raha ho? 👉 Samajh jao — geometry ka gold mine mil chuka hai 🔥 Yahan formula kam , logic zyada kaam karta hai. 🖼️ Concept Image ✍️ Short Explanation Is type ke questions JEE mein isliye favourite hain kyunki: ✔ Algebra aur geometry perfectly combine hote hain ✔ Ek right-angle condition poora problem lock kar deta hai Agar steps clear hain, toh answer clean aur fast nikalta hai. 🔹 Step 1 — Standard Hyperbola Basics (FOUNDATION 💯)** Standard form: x 2 a 2 − y 2 b 2 = 1 Is hyperbola ke liye: c 2 = a 2 + b 2 c^2=a^2+b^2 Foci = ( ± c , 0 ) 📌 Agar focus diya ho , toh c directly mil jaata hai — aur yahin se poora question start hota hai. 🔹 Step 2 — Latus Rectum of Hyperbola Hyperbola ka latus rectum : Har focus se guzarta hai Transverse axis ke perpendicular hota hai Equation: x = ± c x=\pm c Endpoints: ( ±...

  ❓ 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 — ...