If the lines given by $\left(x^2+y^2 ight) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2$ are perpendicular to each other, then $\sin ^2 \alpha+ an ^2 \alpha=$

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If the lines given by $\left(x^2+y^2ight) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2$ are perpendicular to each other, then $\sin ^2 \alpha+	an ^2 \alpha=$

Aicharya · Accepted Answer

Given pair of lines $ \begin{aligned} & \left(x^2+y^2ight) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2 \ & \Rightarrow x^2 \cos 2 \alpha+y^2(0)-2 x y \sin \alpha \cos \alpha=0 \end{aligned} $ Since lines are perpendicular. $ \begin{aligned} & 	herefore 	ext { Coefficient of } x^2+	ext { coefficient of } y^2=0 \ & \Rightarrow \cos 2 \alpha=0 \end{aligned} $ $\begin{aligned} & \Rightarrow 1-2 \sin ^2 \alpha=0 \ & \Rightarrow \sin ^2 \alpha=\frac{1}{2} \ & 	ext { and } \cos ^2 \alpha=1-\frac{1}{2}=\frac{1}{2} \ & 	herefore 	an ^2 \alpha=\frac{\left(\frac{1}{2}ight)}{\left(\frac{1}{2}ight)}=1 \ & 	herefore \sin ^2 \alpha+	an ^2 \alpha=1+\frac{1}{2}=\frac{3}{2}\end{aligned}$

If the lines given by $\left(x^2+y^2\right) \sin ^2 \alpha=(x \cos \alpha-y \sin \alpha)^2$ are…

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