TS EAMCET 14-Sep-2020 Shift 1 Solved Paper

Given curves ax2+by2=1...(i)
and ‌‌cx2+dy2=1....(ii)
From Eqs. (i) and (ii), (a−c)x2+(b−d)y2=0
‌

x2

y2

=−(‌

b−d

a−c

)...(iii)
Differentiating Eq. (i) w.r.t.' x′, we get
2ax+2byy′=0⇒y′=−‌

ax

by

Differentiating Eq. (ii) w.r.t' x′ we get
2cx+2dyy′=0⇒y′=‌

−αx

dy

Curves intersect orthogonally

From Eqs. (iii) and (iv), we get
−(‌

b−d

a−c

)=−‌

bd

ac

−(‌

b−d

a−c

)=−‌

bd

ac