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. $noneofthese$

Answer 1

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

for line $a{x}^2+2hxy+b{y}^2=0$

angle between two lines is given by

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$\begin{array}{cc}\text{for}{x}^2-& \text{2pxy}+y^2=0\\ \tan \theta & =\frac{2\sqrt{p^2-1}}{2}\\ & =\sqrt{p^2-1}\end{array}$

$\Rightarrow {\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow {\sec }^2\theta -1+{\tan }^2\theta$

$\Rightarrow \sec \theta =\sqrt{1+p^2-1}$

$\Rightarrow \sec \theta =p$

$\therefore \theta ={\sec }^{-1}\left(p\right)$

option $A$ is correct.