Question

Find the angle between the lines represented by the equation $x^2-2pxy+y^2=0$

• A. ${\sec }^{-1}\left(p\right)$
• B. ${\tan }^{-1}\left(p\right)$
• C. ${\cos }^{-1}\left(p\right)$
• D. none of these

Answer 1

The correct option is B ${\tan }^{-1}\left(p\right)$

$x^2-2pxy+y^2=0$

$\theta ={\tan }^{-1}\left[\frac{2\sqrt{h^2-ab}}{a+b}\right]$

Where, $\theta$ is the angle between the pair of straight lines.

standard equation $=a{x}^2+2hxy+b{y}^2=0$

$\begin{array}{cc}\therefore a& =1\\ & h=-p\\ & b=1\\ \theta & ={\tan }^{-1}\left[\frac{x\sqrt{p^2-1}}{x}\right]\end{array}$

$\theta ={\tan }^{-1}\left[\sqrt{p^2-1}\right]$

option $A=\theta ={\tan }^{-1}\left[\sqrt{p^2-1}\right]$