Question

If the conics whose equations are
$S_1:\left({\sin }^2θ\right)x^2+(2h\tan θ)xy+\left({\cos }^2θ\right)y^2$
$+32x+16y+19=0$
$S_2:\left({\cos }^2θ\right)x^2−\left(2h^{′}\cot θ\right)xy+\left({\sin }^2θ\right)y^2$
$+16x+32y+19=0$
intersect in four concyclic points, where $θ∈\left(0,\frac{π}{2}\right)$, then
the incorrect statement(s) can be

• A. $h+h^{′}=0$
• B. $h−h^{′}=0$
• C. $θ=\frac{Ï€}{4}$
• D. None of these

Answer 1

Answer: (A)

Curve through the intersection of $S_1$ and $S_2$ is given by $S_1+λS_2=0$
$⇒x^2\left({\sin }^2θ+λ{\cos }^2θ\right)+2\left(h\tan θ−λh^{′}\cot θ\right)xy$
$+\left({\cos }^2θ+λ{\sin }^2θ\right)y^2+(32+16λ)x+(16+32λ)y+19(1+λ)=0$
The above equation will represent a circle if
${\sin }^2θ+λ{\cos }^2θ={\cos }^2θ+λ{\sin }^2θ$
$⇒(1−λ)\left({\sin }^2θ−{\cos }^2θ\right)=0$
$⇒λ=1\text{ or }θ=\frac{π}{4}$
Also $h\tan θ−λh^{′}\cot θ=0$
$⇒h\tan θ=λh^{′}\cot θ\text{ which is satisfied if }λ=1$ and $θ=\frac{π}{4}$
$⇒h=h^{′}$