Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Concept ⭐ Parametric Trigonometric Curves in Coordinate Geometry Trig Parametric Curves in 59 Sec Whenever you see x = f ( θ ) , y = g ( θ ) x = f(\theta),\quad y = g(\theta) samajh jao — curve directly nahi diya , balki parameter ke through hide karke diya gaya hai . 1️⃣ Parametric Form — Actual Meaning In parametric equations: x x  and y y  are written in terms of a parameter (usually θ \theta ) As θ \theta  varies → point ( x , y ) (x,y)   moves The path traced by this moving point is the required curve 👉 Direct graph nahi, motion-based definition hoti hai. 2️⃣ Why tan(θ + constant) is Used In JEE, parametric trig curves often look like: x = a tan ⁡ ( θ + A ) , y = b tan ⁡ ( θ + B ) x = a\tan(\theta + A),\quad y = b\tan(\theta + B) Reason 👇 tan-shift identities relate angles linearly Difference of angles becomes constant This allows easy elimination of θ \theta 📌 Especially powerful when: ( θ + A ) − ( θ + B ) = consta...

❓ Question:  Find the area of the region { ( x , y ) :    1 + x 2 ≤ y ≤ min ⁡ {   x + 7 ,    11 − 3 x   } } . If this area is A A , compute 3 A 3A . 🖼️ Question Image ✍️ Short Solution Find where the two lines cross each other. Compare x + 7 x+7  and 11 − 3 x 11-3x : x + 7 ≤ 11 − 3 x    ⟺    4 x ≤ 4    ⟺    x ≤ 1. So on ( − ∞ , 1 ] (-\infty,1]  the upper boundary is x + 7 x+7 ; on [ 1 , ∞ ) [1,\infty)  the upper boundary is 11 − 3 x 11-3x . Both meet at x = 1 x=1  with value 8 8 . Find intersection points of parabola with each line. With y = x + 7 y=x+7 : solve 1 + x 2 = x + 7 ⇒ x 2 − x − 6 = 0 1+x^{2}=x+7 \Rightarrow x^{2}-x-6=0  → x = − 2 ,   3 x=-2,\,3 . With y = 11 − 3 x y=11-3x : solve 1 + x 2 = 11 − 3 x ⇒ x 2 + 3 x − 10 = 0 →  x = − 5 ,   2 x=-5,\,2 . Determine the x-range where the parabola lies below the relevant line. For the branch where upper = x + 7 x+7  (valid for x ≤ 1 x\le1 ), the parabola is below this...