Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Parabola: Focus–Directrix Trick Parabola ka equation yaad nahi? 👉 Koi tension nahi! Bas focus–directrix definition yaad rakho — aur poora parabola khud-ba-khud ban jaata hai. 🖼️ Concept Image ✍️ Short Explanation Parabola ek distance-based curve hai. Equation derivation ya formula yaad karne ki zarurat hi nahi — sirf distance equality lagao, square karo, equation mil jaata hai. 🔹 Step 1 — Definition of Parabola Parabola = locus of a point P(x, y) such that Distance(P, Focus) = Distance(P, Directrix) Bas — yeh hi official definition hai 🔥 🔹 Step 2 — Distance Formula Concept ✔ Distance from focus → point-to-point distance using ( x − x 1 ) 2 + ( y − y 1 ) 2​ ✔ Distance from directrix → perpendicular distance from a line ∣ a x + b y + c ∣ a 2 + b 2​ 👉 In dono ko equal rakhte hain. 🔹 Step 3 — General Method (Always Works 💯)** Let P ( x , y ) P(x,y)  lie on parabola. Write: Distance(P, Focus) = Distance(P,...

❓ Question If the shortest distance between the lines x − 1 2 = y − 2 3 = z − 3 4 and x 1 = y α = z − 5 1 \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad\text{and}\quad \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} is 5 6 , then the sum of all possible values of α \alpha α is equal to ? 🖼️ Question Image ✍️ Short Solution We’ll use the vector formula for the shortest distance between two skew lines : SD = ∣ ( b ⃗ − a ⃗ ) ⋅ ( d 1 ⃗ × d 2 ⃗ ) ∣ ∣ d 1 ⃗ × d 2 ⃗ ∣ where d ⃗ 1 , d ⃗ 2 \vec d_1,\vec d_2 are direction vectors and a ⃗ , b ⃗ \vec a,\vec b  are position vectors of any points on the lines. 🔹 Step 1 — Identify data from line equations Line 1 x − 1 2 = y − 2 3 = z − 3 4 \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} Point A ( 1 , 2 , 3 ) , d ⃗ 1 = ( 2 , 3 , 4 ) A(1,2,3),\quad \vec d_1=(2,3,4) Line 2 x 1 = y α = z − 5 1 \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} Point (when parameter = 0): B ( 0 , 0 , 5 ) , d ⃗ 2 = ( 1 , α , 1 ) B(0,0,5),\quad \vec d_2=(1,\alpha...

  ❓ Question From a group of 7 batsmen and 6 bowlers , 10 players are to be chosen for a team, which should include: ✔ at least 4 batsmen ✔ at least 4 bowlers ✔ and MUST include one batsman (captain) and one bowler (vice-captain) Find the total number of ways such a team can be selected. 🖼️ Question Image ✍️ Short Solution This is a classic combinations + constraints problem. 👉 Since captain (batsman) and vice-captain (bowler) must be selected , we include them first — then count valid ways to choose the remaining players abiding by the minimum batsman–bowler rule. 🔹 Step 1 — Fix the compulsory players We must include: 1 batsman (captain) 1 bowler (vice-captain) So: Players already selected = 2 \text{Players already selected} = 2 Remaining to choose: 10 − 2 = 8 10 - 2 = 8 🔹 Step 2 — Update available players After fixing the captain & vice-captain: Remaining batsmen = 7 − 1 = 6 7 - 1 = 6 Remaining bowlers = 6 − 1 = 5 6 - 1 = ...

  ❓ Concept 🎬 Remainder Tricks for Huge Powers Kabhi-kabhi JEE mein aise questions milte hain jahan power itna bada hota hai ki calculator bhi give-up kar de 😅 Tab kaam aata hai modulo + cyclic pattern — aur poora question 1-min trick se solve ho jaata hai. 🖼️ Concept Image ✍️ Short Explanation Basic funda: 👉 Remainder depends on modulo behaviour — not on size of the number. Isliye pehle number ko chhota banao, phir power handle karo. 🔹 Step 1 — Remainder Concept If a ≡ r ( m o d n ) a \equiv r \pmod n then a k ≡ r k ( m o d n ) a^k \equiv r^k \pmod n 📌 Matlab: Pehle base ko modulo se reduce karo Phir power lagao 🔹 Step 2 — Reduce the Base First (Always!) Large number ko directly power nahi karte. Pehle: a m o d     n = r a \mod n = r Phir sirf r ke saath kaam karo — calculation ultra-simple ho jaata hai. Example: 388 64 m o d     7 ⇒ 388 ≡ 3 m o d     7 388^{64} \mod 7 \Rightarrow 388 \equiv 3 \mod 7 So: 388 64 ≡ 3 64 m o d     7 38...

  ❓ Concept 🎬 3D Line Through a Point & Two Lines JEE 3D geometry mein ek super-common pattern hota hai: 👉 A line L passes through a fixed point and intersects two given lines. Iska matlab hai — L ko define karne ke liye bas parameter aur collinearity ka game khelo. ✍️ Short Explanation Trick simple hai: ✔ Dono lines ki parametric form likho ✔ Assume intersection points ✔ Use collinearity with the fixed point ✔ Solve parameters ✔ Get direction of L → equation ready! 🔹 Step 1 — Write Given Lines in Parametric Form Suppose: r ⃗ = a ⃗ 1 + λ d ⃗ 1 \vec r = \vec a_1 + \lambda \vec d_1 r ⃗ = a ⃗ 2 + μ d ⃗ 2 \vec r = \vec a_2 + \mu \vec d_2 Where a ⃗ 1 ,   a ⃗ 2 \vec a_1,\ \vec a_2  = position vectors of known points d ⃗ 1 ,   d ⃗ 2 \vec d_1,\ \vec d_2 ​ = direction vectors 🔹 Step 2 — Find Intersection Points with L Assume the line L passes through a fixed point P P P . Let A = a ⃗ 1 + λ d ⃗ 1 A = \vec a_1 + \lambda \vec d_1 B = ...

  ❓ Question Let the system of equations { 2 x + 3 y + 5 z = 9 7 x + 3 y − 2 z = 8 12 x + 3 y − ( 4 + λ ) z = 16 − μ \begin{cases} 2x + 3y + 5z = 9 \\ 7x + 3y - 2z = 8 \\ 12x + 3y - (4 + \lambda)z = 16 - \mu \end{cases} have infinitely many solutions . Then the radius of the circle centred at ( λ , μ ) (\lambda,\mu)  and touching the line 4 x = 3 y 4x = 3y is equal to ? 🖼️ Question Image ✍️ Short Solution This question beautifully mixes: ✔ Condition for infinitely many solutions in 3 variables ✔ Linear dependence of equations ✔ Distance of a point from a line (circle tangent condition) Let’s crack it step-by-step. 🔹 Step 1 — Infinite solutions condition A system of 3 linear equations in 3 variables has infinitely many solutions 👉 iff the third equation is a linear combination of the first two (so rank < 3). So we assume ( 3 ) = a ( 1 ) + b ( 2 ) (3) = a(1) + b(2) That is: 12 x + 3 y − ( 4 + λ ) z = 16 − μ 12x + 3y - (4+\lambda)z = 16-\mu must eq...

  ❓ Concept 🎬 Exact DE Trick JEE mein jab differential equation dikhe: M ( x , y )   d y + N ( x , y )   d x = 0 sabse pehla sawaal hona chahiye 👇 👉 “Kya yeh exact hai?” Agar exact nikla — solution almost direct mil jaata hai . 🖼️ Concept Image ✍️ Short Explanation Exact DE questions JEE ke sabse scoring problems hote hain. Reason simple hai: 👉 Na substitution 👉 Na heavy algebra 👉 Sirf partial derivatives + integration 🔹 Step 1 — Form of an Exact Differential Equation Given form: M ( x , y )   d y + N ( x , y )   d x = 0 M(x,y)\,dy + N(x,y)\,dx = 0 ✅ Exactness Condition: ∂ M ∂ x = ∂ N ∂ y 📌 Agar dono equal aa jaayein: DE exact hai Direct integration allowed Agar equal nahi: Phir integrating factor ka case hota hai (alag topic) 🔹 Step 2 — Meaning of Exactness Exact hone ka matlab 👇 Ek function F ( x , y ) F(x,y)  exist karta hai aisa ki: d F = M   d y + N   d x dF = M\,dy + N\,dx Isliye DE ka general solution hota hai: F ( x...