TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Equation of line is x+y=1 . . . (i)
and equation of the curve is
x2+y2+2hxy+gx+fy+1=0 . . . (ii)
Making Eq. (ii) homogeneous by Eq. (i), we get the equation of the lines joining the origin to the point of intersection of Eqs. (i) and (ii).
x2+y2+2hxy+(gx+fy)‌(x+y)+1(x+y)2=0
⇒x2+y2+2hxy+gx2+gxy
‌+fxy+fy2+x2+y2+2xy=0
⇒(2+g)x2+(2+f)y2+xy(g+f+2h+2)=0‌‌...‌ (iii) ‌
Lines denoted by Eq. (iii) will be perpendicular to each other if
Coefficient of x2+ Coefficient of y2=0
i.e. (2+g)+(2+f)=0⇒g+f+4=0
Locus of (g,f) is x+y+4=0
i.e. (g,f) lies on x+y=−4.