A) \[{{\cos }^{-1}}(p)\]
B) \[{{\sec }^{-1}}(p)\]
C) \[{{\sec }^{-1}}(-p)\]
D) \[{{\sec }^{-1}}(\pm p)\]
Correct Answer: B
Solution :
Let \[\theta \] be the angle between the lines represented by \[{{x}^{2}}-2pxy+{{y}^{2}}=0\] Comparing the given equation with \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] We get,\[a=1,\,\,b=1\]and\[2h=-2p\] \[\therefore \] \[\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] \[\Rightarrow \] \[\tan \theta =\frac{2\sqrt{{{p}^{2}}-1}}{1+1}=\sqrt{{{p}^{2}}-1}\] \[\Rightarrow \] \[{{p}^{2}}=1+{{\tan }^{2}}\theta \] \[\Rightarrow \] \[{{p}^{2}}={{\sec }^{2}}\theta \] \[\Rightarrow \] \[\sec \theta =\pm p\Rightarrow \theta ={{\sec }^{-1}}(\pm p)\] Hence, the required acute angle is\[{{\sec }^{-1}}(p)\].
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