Question

If the conics whose equations are $S \equiv x ^2 \sin ^2 \theta+2 h x y+y^2 \cos ^2 \theta+32 x+16 y+19=0$ & $S ^{\prime} \equiv x ^2 \cos ^2 \theta+2 h^{\prime} xy + y ^2 \sin ^2 \theta+16 x +32 y +19=0$ intersects in four concylic points then

• A. $ h+h^{\prime}=0$
• B. $h=h^{\prime}$
• C. $h+h^{\prime}=1$
• D. $h+h^{\prime}=2$

Answer 1

Answer: (A)

$S+\lambda S^{\prime}=0$
$\Rightarrow x^2\left(\sin ^2 \theta+\lambda \cos ^2 \theta\right)+y^2\left(\cos ^2 \theta+\lambda \sin ^2 \theta\right)+2 x y\left(h+\lambda h^{\prime}\right)+x(32+16 \lambda)+y(16+32 \lambda)+19(1+\lambda)=0$
it will represent a circle it
$\sin ^2 \theta+\lambda \cos ^2 \theta=\cos ^2 \theta+\lambda \sin ^2 \theta \& h+\lambda h^{\prime}=0$
$\lambda=1 \therefore h+h^{\prime}=0$