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If pairs of straight lines \(x^2-2 p x y-y^2=0\) and \(x^2-2 q x y-y^2=0\) be such that each pair bisects the angle between the other pair, then
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Verified Answer
The correct answer is:
\(p q=-1\)
Equation of angle bisector for
\(\begin{array}{rlrl}
a x^2+2 h x y+b y^2 & =0 \\
\Rightarrow & \frac{x^2-y^2}{a-b} & =\frac{x y}{h}
\end{array}\)
and for \(x^2-2 p x y-y^2=0\)
\(\Rightarrow \quad \frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p} \Rightarrow x^2-y^2+\frac{2 x y}{p}=0\)
Given equation of angle bisector is
\(x^2-2 q x y+y^2=0\)
On comparing
\(\Rightarrow \quad \frac{2}{P}=-2 q \Rightarrow p q=-1\)
\(\begin{array}{rlrl}
a x^2+2 h x y+b y^2 & =0 \\
\Rightarrow & \frac{x^2-y^2}{a-b} & =\frac{x y}{h}
\end{array}\)
and for \(x^2-2 p x y-y^2=0\)
\(\Rightarrow \quad \frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p} \Rightarrow x^2-y^2+\frac{2 x y}{p}=0\)
Given equation of angle bisector is
\(x^2-2 q x y+y^2=0\)
On comparing
\(\Rightarrow \quad \frac{2}{P}=-2 q \Rightarrow p q=-1\)
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