Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question Let P P  be the parabola whose focus is ( − 2 ,   1 ) (-2,\,1) and whose directrix is 2 x + y + 2 = 0. 2x + y + 2 = 0. Find the sum of the ordinates of the points on P P P whose abscissa is x = − 2. 🖼️ Question Image ✍️ Short Solution A parabola is defined as the locus of a point whose distance from the focus equals its perpendicular distance from the directrix . We will: 1️⃣ Write the focus–directrix condition 2️⃣ Substitute x = − 2 x = -2 3️⃣ Solve for y y 4️⃣ Add the ordinates 🔹 Step 1 — Use focus–directrix definition Let ( x , y ) (x,y)  be any point on the parabola. Distance from focus ( − 2 , 1 ) (-2,1) : ( x + 2 ) 2 + ( y − 1 ) 2 \sqrt{(x+2)^2 + (y-1)^2} ​ Distance from directrix 2 x + y + 2 = 0 2x + y + 2 = 0 : ∣ 2 x + y + 2 ∣ 2 2 + 1 2 = ∣ 2 x + y + 2 ∣ 5 \frac{|2x + y + 2|}{\sqrt{2^2 + 1^2}} = \frac{|2x + y + 2|}{\sqrt{5}} ​ Equating squares: ( x + 2 ) 2 + ( y − 1 ) 2 = ( 2 x + y + 2 ) 2 5 (x+2)^2 + (y-1)^2 = \frac{...

  ❓ Question FOR: Let C 1 C_1 ​ be the circle in the third quadrant of radius 3 , that touches both coordinate axes . Let C 2 C_2  be the circle with centre ( 1 ,   3 ) (1,\,3)  that touches C 1 C_1   externally at the point ( α ,   β ) (\alpha,\,\beta) . If ( β − α ) 2 = m n , gcd ⁡ ( m , n ) = 1 , (\beta - \alpha)^2 = \frac{m}{n}, \quad \gcd(m,n)=1, then the value of m + n m+n is equal to ? 🖼️ Question Image ✍️ Short Solution This problem uses pure coordinate geometry logic : 👉 A circle touching both axes has its centre fixed by symmetry. 👉 When two circles touch externally , the point of contact lies on the line joining their centres . 👉 We find that point using section formula , then compute ( β − α ) 2 (\beta-\alpha)^2 . 🔹 Step 1 — Equation and centre of C 1 C_1 ​ Circle C 1 C_1 C 1 ​ : Lies in third quadrant Touches x-axis and y-axis Radius = 3 Hence, its centre must be: ( − 3 ,   − 3 ) (-3,\,-3) (Only this point keeps t...

  ❓ 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...