TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

We have,
x2+y2=a2 . . . (i)
and xy=b2 . . . (ii)Let four points
A(x1,y1),B(x2,y2),C(x3,y3)
and D(x4,y4)
∴‌‌OA2=x12+y12
OB2=x22+y22
OC2=x32+y32
and ‌‌OD2=x42+y42
Such that OA2+OB2+OC2+OD2
‌‌=Σx12+Σy12=4a2
From Eq. (ii) x=‌

b2

y

From Eq. (i) (‌

b2

y

)‌

y

2

+y2=a2
⇒‌‌b4+y4=a2y2 ⇒‌‌y4−a2y2+b4=0
This is an equation of 4 th degree in y and its four roots are y1,y2,y3,y4, then y1y2y3y4=b4