Question

If $(a+3b)(3a+b)=4h^2$then the angle between the lines represented by $a{x}^2+2hxy+b{y}^2=0$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${\tan }^{-1}(½)$

Answer 1

The correct option is C

${60}^0$

The explanation for the correct option:

Finding the angle:

Given $(a+3b)(3a+b)=4h^2$

$$\begin{gathered} \Rightarrow 3a^2+9ab+ab+3b^2=4h^2 \\ \Rightarrow 3a^2+10ab+3b^2=4h^2..\left(i\right) \end{gathered}$$

We know the angle between pair of straight lines $a{x}^2+2hxy+b{y}^2=0$

$$\begin{gathered} \tan \theta =2\frac{\sqrt{(h^2–ab)}}{(a+b)} \\ =\frac{\sqrt{(4h^2–4ab)}}{(a+b)} \\ =\frac{\sqrt{(3a^2+10ab+3b^2–4ab)}}{(a+b)}------(from(i\left)\right) \\ =\frac{\sqrt{(3a^2+6ab+3b^2)}}{(a+b)} \\ =\frac{\sqrt{3(a^2+2ab+b^2)}}{(a+b)} \\ =\frac{\sqrt{3{(a+b)}^2}}{(a+b)} \\ =\frac{\sqrt{3}(a+b)}{(a+b)} \\ =\sqrt{3} \end{gathered}$$

$\Rightarrow \theta ={\tan }^{-1}\left(\sqrt{3}\right)={60}^0$

Hence, option (C) is the correct answer.