xy=1,−1
2t1+t2⋅2t11−t21=1
⇒t22−t1=24t1t2
Product of slope =−1
t121×(−t221)=−1⇒t1t2=1
⇒(t1t2)2=1⇒t1t2=1
⇒t12+t22=42+4=25
⇒t12=2+5⇒t121=5−2
AB2=(t1−t2)2+(t11+t21)2
=2(t12+t121)=45⇒ Area 2=80 sq. unit
A square ABCD has all its vertices on the curve x2y2=1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is
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