JEE Main 10 Jan 2019 Shift 2 Solved Paper

x2 = 4y
x - √2 y + 4√2 = 0
Solving together we get
x2 = 4 (

x+4√2

√2

)
√2x2 + 4x + 16√2
√2x2 - 4x - 16√2 = 0
x1+x2 = 2√2 ; x1x2 =

−16√2

√2

= - 16
Similarly,
(√2y−4√2)2 = 4y
2y2 + 32 - 16y = 4y
2y2 - 20y + 32 = 0 → y1+y2 = 10 , y1y2 = 16

lAB = √(x2−x1)2+(y2−y1)2
= √(2√2)2+64+(10)2−4(16)
= √8+64+100−64
= √108 = 6 √3