MHT CET · Maths · Pair of Lines
If \(x^{2}-2 p x y-y^{2}=0\) and \(x^{2}-2 q x y-y^{2}=0\)
bisect angles between each other, then
- A \(p+q=1\)
- B \(p q=1\)
- C \(p q+1=0\)
- D \(p^{2}+p q+q^{2}=0\)
Answer & Solution
Correct Answer
(C) \(p q+1=0\)
Step-by-step Solution
Detailed explanation
The equation of the bisectors of the angles between the lines \(x^{2}-2 p x y-y^{2}=0\) is
\(\frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\)
\(\Rightarrow \frac{x^{2}-y^{2}}{2}=\frac{-x y}{p}\)
\(\Rightarrow p x^{2}+2 x y-p y^{2}=0\)
This is same as \(x^{2}-2 q x y-y^{2}=0\). Therefore
\(
\frac{p}{1}=\frac{2}{-2 q}=-\frac{p}{-1}
\)
\(
\Rightarrow \quad p q+1=0
\)
\(\frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\)
\(\Rightarrow \frac{x^{2}-y^{2}}{2}=\frac{-x y}{p}\)
\(\Rightarrow p x^{2}+2 x y-p y^{2}=0\)
This is same as \(x^{2}-2 q x y-y^{2}=0\). Therefore
\(
\frac{p}{1}=\frac{2}{-2 q}=-\frac{p}{-1}
\)
\(
\Rightarrow \quad p q+1=0
\)
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