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  • If the conics whose equation are <img alt="\mathrm{S \equiv \sin ^2 \theta x^2+2 h x y+\cos ^2 \theta y^2+32 x+16y+19=0}" src="https://entrancecorner.oncodecogs.com/

If the conics whose equation are \mathrm{S \equiv \sin ^2 \theta x^2+2 h x y+\cos ^2 \theta y^2+32 x+16y+19=0} 

\mathrm{S^{\prime} \equiv \cos ^2 \theta x^2+2 h^{\prime} x y+\sin ^2 \theta y^2+16 x+32 y+19=0} intersects in four concyclic points then, (where θ ∈ R)

Option: 1

h + h′ = 0


Option: 2

h = h′ 

 


Option: 3

h + h′ = 1


Option: 4

none of these

 


Answers (1)

best_answer

Curve passing through point of intersection of S and S′ is 

\mathrm{\begin{aligned} & \Rightarrow S+\lambda S^{\prime}=0 \\ & \Rightarrow x^2\left(\sin ^2 \theta+\lambda \cos ^2 \theta\right)+y^2\left(\cos ^2 \theta+\lambda \sin ^2 \theta\right)+2 x y\left(h+\lambda h^{\prime}\right)+x(32+16 \lambda) \\ & +y(16+32 \lambda)+19(1+\lambda)=0 \end{aligned}}

for this equation to be a circle 

\mathrm{\begin{aligned} & \sin ^2 \theta+\lambda \cos ^2 \theta=\cos ^2 \theta+\lambda \sin ^2 \theta \Rightarrow \lambda=1 \\ & \text { and } h+\lambda h^{\prime}=0 \Rightarrow h+h^{\prime}=0 . \end{aligned}}

 

 

Posted by

Rakesh

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