If the conics whose equations are
S 1 : (sin 2 θ ) x 2 + (2h tan θ )xy + (cos 2 θ ) y 2 + 32x + 16y + 19 = 0
S 2 : (cos 2 θ ) x 2 – (2h' cot θ )xy + (sin 2 θ ) y 2 + 16x + 32y + 19
= 0
intersect in four concyclic points, where θ ∈ 
, then the correct statement (s) can be-
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(a, b, c)
Curve through the intersection of S 1 and S 2 is given by
S 1 + λ S 2 = 0
⇒ x 2 (sin 2 θ + λ cos 2 θ ) + 2(h tan θ + λ h' cot θ ) xy –
(cos 2 θ + λ sin 2 θ )y 2 + (32 + 16 λ ) x + (16 + 32 λ ) y + 19(1 + λ ) = 0
The above equation will represent a circle if
sin 2 θ + λ cos 2 θ = cos 2 θ + λ sin 2 θ
⇒ sin 2 θ + λ sin 2 θ = cos 2 θ + λ cos 2 θ + λ – λ cos 2 θ
⇒ (1 – λ ) sin 2 θ = (1 – λ )cos 2 θ
⇒ (1 – λ ) (sin 2 θ – cos 2 θ ) = 0
λ = 1 or θ = 
htan θ – λ h'cot θ = 0
⇒ htan θ = λ h'cot θ which is satisfied if
λ = 1 and θ = 
⇒ h = h' or λ = – 1, θ = 
⇒ h + h' = 0
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