Question

The absolute value of difference of the slope of the lines $x^2({\sec }^2\theta -{\sin }^2\theta )-2xy\tan \theta +y^2{\sin }^2\theta =0$ is

• A. $-2$
• B. $\frac{1}{2}$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Given pair of lines

$x^2({\sec }^2\theta -{\sin }^2\theta )-2xy\tan \theta +y^2{\sin }^2\theta =0$

$y^2{\sin }^2\theta -2xy\tan \theta +x^2({\sec }^2\theta -{\sin }^2\theta )=0$

$y=\frac{2x\tan \theta \pm \sqrt{4x^2{\tan }^2\theta -4{\sin }^2\theta x^2({\sec }^2\theta -{\sin }^2\theta )}}{2{\sin }^2\theta }$

$y=\frac{2x\tan \theta \pm \sqrt{4x^2{\tan }^2\theta -4{\tan }^2\theta x^2+4x^2{\sin }^4\theta }}{2{\sin }^2\theta }$

$y=\frac{2x\tan \theta \pm \sqrt{4x^2{\sin }^4\theta }}{2{\sin }^2\theta }$

$y=\frac{2x\tan \theta \pm 2x{\sin }^2\theta }{2{\sin }^2\theta }$

$y=\frac{(\tan \theta \pm {\sin }^2\theta )}{{\sin }^2\theta }x$

On comparing above equation with $y=mx+c$ we get

$m=\frac{(\tan \theta \pm {\sin }^2\theta )}{{\sin }^2\theta }$

Here $m_1=\frac{(\tan \theta +{\sin }^2\theta )}{{\sin }^2\theta }$ and $m_2=\frac{(\tan \theta -{\sin }^2\theta )}{{\sin }^2\theta }$

$m_1-m_2=\frac{(\tan \theta +{\sin }^2\theta )}{{\sin }^2\theta }-\frac{(\tan \theta -{\sin }^2\theta )}{{\sin }^2\theta }$

$\Rightarrow m_1-m_2=\frac{\tan \theta +{\sin }^2\theta -\tan \theta +{\sin }^2\theta }{{\sin }^2\theta }$

$\Rightarrow m_1-m_2=\frac{2{\sin }^2\theta }{{\sin }^2\theta }$

$\Rightarrow m_1-m_2=2$