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JEE Main20246 Apr 2024Morning ShiftMathematicsParabolaActual

Let L₁, L₂ be the lines passing through the point P(0,1) and touching the parabola 9 x^2+12 x+18 y-14=0 . Let Q and R be the points on the lines L₁ and L₂ such that the P Q R is an isosceles triangle with base Q R . If the slopes of the lines Q R are m₁ and m₂ , then 16 (m₁^2+m₂^2 ) is equal to _______

Correct answer

0

Step-by-step solution

aligned & 9 x^2+12 x+4=-18(y-1) & (3 x+2)^2=-18(y-1) & (x+ 2 3 )^2=-2(y-1) aligned aligned & (0,1) & y=m x+1 & (x+ 2 3 )^2=-2(y-1) & (3 x+2)^2=-18 m x & 9 x^2+(12+18 m) x+4=0 & 4(6+9 m)^2=4(36) & 6+9 m=6,-6 & m=0, -4 3 aligned aligned & =- 4 3 & 2 2 1- ^2 2 = -4 3 & ( 2 -2 ) (2 2 +1 )=0 & 2 =2, -1 2 & ~m _ QR = (90+ 2 ) aligned aligned & =- 2 & ~m ₁= -1 2 ~m ₂= -1 -1 / 2 =2 & 16 ( ~m ₁^2+ m ₂^2 )=16 ( 1 4 +4 ) & =4+64=68 aligned

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