Question

The acute angle between the lines represented by $(x^2+y^2)\sqrt{3}=4xy$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. None of these

Answer 1

The correct option is A

$\frac{\pi }{6}$

The explanation for the correct option:

Step1. Combined equation of the two lines :

Let us have below two lines that represent the given equation.

$y-m_{1}x=0$ and $y-m_{2}x=0$

Now, the Combined equation of these two lines will be,

$(y-m_{1}x)(y-m_{2}x)=0$

$y^2-(m_1+m_2)xy+\left(m_1{m}_2\right)x^2=0......(1)$

Now, the given equation is,

$\Rightarrow (x^2+y^2)\sqrt{3}=4xy$

$\Rightarrow (x^2+y^2)-\frac{4}{\sqrt{3}}xy=0$

Comparing this with the given equation $\left(1\right)$, we get

$m_1+m_2=\frac{4}{\sqrt{3}}$and $m_1{m}_2=1$

Now, we know that

${(m_1-m_2)}^2={(m_1+m_2)}^2-4m_1{m}_2$

$\Rightarrow$ ${(m_1-m_2)}^2=\frac{16}{3}-4$

$\Rightarrow$ ${(m_1-m_2)}^2=\frac{4}{3}$

$\Rightarrow$ $|m_1-m_2|=\frac{2}{\sqrt{3}}$

Step2. Find the acute angle $\alpha$:

Now, for the acute angle $\alpha$ between these two lines,

we have

$\tan \alpha =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow$$\tan \alpha =\left|\frac{{\displaystyle \frac{2}{\sqrt{3}}}}{1+1}\right|$

$\Rightarrow$$\tan \alpha =\left|\frac{1}{\sqrt{3}}\right|$

$\Rightarrow$$\tan \alpha =\tan \left(\frac{\pi }{6}\right)$

$\Rightarrow$ $\alpha =\frac{\pi }{6}$

Hence, the correct option is (A).