Question

The equations of the lines $L_1$ and $L_2$ are $y=mx$ and $y=nx,$ respectively. Suppose $L_1$ makes twice as large an angle with the horizontal (measured counterclockwise from the positive $x$ -axis) as does $L_2$ and $m=4n$ , then the value of $\frac{\left(m^2+4n^2\right)}{\left(m^2-6n^2\right)}$ is equal to (where, $n\ne 0$ )

• A. $3$
• B. $-3$
• C. $2$
• D. $-2$

Answer 1

Answer: (C)

Let, $m=tan2\theta$ and $n=tan\theta$
And $m=4n$ and $m\ne 0,$
$\therefore tan2\theta \ne 0;\theta \ne 0$
$tan2\theta =4tan\theta$
$\frac{2tan\theta }{1-ta{n}^2\theta }=4tan\theta$
$\therefore 1=2-2ta{n}^2\theta \Rightarrow 2ta{n}^2\theta =1$
$\Rightarrow ta{n}^2\theta =\frac{1}{2}=n^2$
$\Rightarrow m^2=16n^2=8\Rightarrow \frac{m^2+4n^2}{m^2-6n^2}=\frac{10}{5}=2$