Question

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

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

Answer 1

The correct option is A

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

Explanation for Correct Answer:

Given the equation $x^2-2pxy+y^2=0$

Now comparing the given equation with the general form of an equation of pair of a lines $a{x}^2+2hxy+b{y}^2=0$

On comparing we get, $a=1$,$b=1$,$2h=-2p\mathrm{or}h=-p$

We know that the angle between the two lines is given by:

$\Rightarrow \tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|...\left(1\right)$

Substituting the values $a=1$ ,$b=1$ and $h=-p$ in the equation $\left(1\right)$, we get

$$\begin{gathered} \tan \theta =\left|\frac{2\sqrt{{\left(-p\right)}^2-\left(1\right)\left(1\right)}}{1+1}\right| \\ =\left|\frac{2\sqrt{p^2-1}}{2}\right| \\ =\left|\sqrt{p^2-1}\right| \end{gathered}$$

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

Now, squaring on both sides

$$\begin{gathered} {\left(\tan \theta \right)}^2={\left(\left|\sqrt{p^2-1}\right|\right)}^2 \\ =p^2-1 \end{gathered}$$

Adding $1$ on both sides, we have

$$\begin{gathered} 1+{\tan }^2\theta =p^2-1+1 \\ =p^2 \end{gathered}$$

Since, ${\sec }^2\left(\theta \right)=1+{\tan }^2\theta$, substituting the value we have,

${\sec }^2\left(\theta \right)=p^2$

By taking square root on both sides, We get,

$$\begin{gathered} \Rightarrow \sec \left(\theta \right)=p \\ \Rightarrow \theta ={\sec }^{-1}\left(p\right) \end{gathered}$$

Therefore, the angle between the two lines is ${\sec }^{-1}\left(p\right)$

Hence, option (A) is correct.