TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Given equation of line x−y=0
Distance from P(α,β) to line x−y=0 is
b=|‌

α−β

√2

|⇒(α−β)2=2b2 . . . (i)
Distance between P(α,β) and A(2,1) is
a2=(α−2)2+(β−1)2 . . . (ii)
Distance from origin to A(2.1) is C2=5
Alsoa=bc
∴‌‌a2=5b2 . . . (iii)
From Eqs. (i), (ii) and (iii), we get
(α−2)2+(β−1)2=‌

5(α−β)2

2

2(α2−4α+4+β2−2β+1]=5[α2+β2−2αβ]
⇒3(α2+β2)−10αβ+8α+4β−10=0
∴ Locus of P is 3(x2+y2)−10xy+8x+4y−10=0