JEE Main 18 March 2021 Shift 1 Solved Paper

Let A(t1,‌

1

t1

) and C(t3,‌

1

t3

) lies on xy=1 and B(t2,‌

1

t2

) and
D(t4,‌

1

t4

) lies on xy=−1 as shown in figure.
ABCD is a square and by symmetry, diagonals must intersects at O(0,0).
⇒‌‌‌

−1

t12

⋅‌

1

t22

=−1
⇒t12t22=1....(i)
Alss, mid-point of AB is P(‌

t1+t2

2

,‌

t2−t1

2t1t2

) lies on xy=1 ⇒‌‌(‌

t1+t2

2

)(‌

t2−t1

2t1t2

)=1
⇒‌‌t22−t12=4t1t2....(ii)
Now, area of square =(AB)2
(t1−t2)2+[‌

1

t1

−(‌

−1

t2

)]2
=(t1−t2)2+(t2+t1)2 [From Eq. (i)]
=2(t12+t22)
=2√20=√80