Question

If the conics whose equations are
$S_{1}:\left(\sin ^{2} \theta\right) x^{2}+(2 h \tan \theta) x y+\left(\cos ^{2} \theta\right) y^{2} $
$+32 x+16 y+19=0 $
$S_{2}:\left(\cos ^{2} \theta\right) x^{2}-\left(2 h^{\prime} \cot \theta\right) x y+\left(\sin ^{2} \theta\right) y^{2} $
$+16 x+32 y+19=0$
intersect in four concyclic points, where $\theta \in\left(0, \frac{\pi}{2}\right)$, then
the incorrect statement(s) can be

• A. $\quad h+h'=0$
• B. $h-h'=0$
• C. $ \theta=\frac{\pi}{4}$
• D. None of these

Answer 1

Answer: (A)

Curve through the intersection of $S_{1}$ and $S_{2}$ is given by $S_{1}+\lambda S_{2}=0$
$\Rightarrow x^{2}\left(\sin ^{2} \theta+\lambda \cos ^{2} \theta\right)+2\left(h \tan \theta-\lambda h'\cot \theta\right) x y$
$+\left(\cos ^{2} \theta+\lambda \sin ^{2} \theta\right) y^{2}+(32+16 \lambda) x+(16+32 \lambda) y+19(1+\lambda)=0$
The above equation will represent a circle if
$ \sin ^{2} \theta+\lambda \cos ^{2} \theta=\cos ^{2} \theta+\lambda \sin ^{2} \theta $
$\Rightarrow (1-\lambda)\left(\sin ^{2} \theta-\cos ^{2} \theta\right)=0 $
$\Rightarrow \lambda=1 \text { or } \theta=\frac{\pi}{4} $
Also $ h \tan \theta-\lambda h^{\prime} \cot \theta=0 $
$\Rightarrow h \tan \theta=\lambda h^{\prime} \cot \theta \text { which is satisfied if } \lambda=1 $ and $\theta=\frac{\pi}{4} $
$\Rightarrow h=h'$