Question

Show that the two straight lines
$x^2({\tan }^2\theta +{\cos }^2\theta )-2xy\tan \theta +y^2{\sin }^2\theta =0$
make with the axis of x angles such that the difference of their tangents is 2.

Answer 1

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

$y=\frac{-(-2x\tan \theta )\pm \sqrt{4x^2{\tan }^2\theta -4\left({\sin }^2\theta \right)x^2({\tan }^2\theta +{\cos }^2\theta )}}{2{\sin }^2\theta }$

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

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

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

$y=\frac{x\tan \theta \pm x\sqrt{\frac{{\sin }^2\theta (1-{\sin }^2\theta -{\cos }^4\theta )}{{\cos }^2\theta }}}{{\sin }^2\theta }=\frac{x\tan \theta \pm x\sqrt{\frac{{\sin }^2\theta ({\cos }^2\theta -{\cos }^4\theta )}{{\cos }^2\theta }}}{{\sin }^2\theta }$

$y=\frac{x\tan \theta \pm x\sqrt{\frac{{\sin }^2\theta {\cos }^2\theta (1-{\cos }^2\theta )}{{\cos }^2\theta }}}{{\sin }^2\theta }=\frac{x\tan \theta \pm x{\sin }^2\theta }{{\sin }^2\theta }=\frac{x\tan \theta }{{\sin }^2\theta }\pm x$

$y=x\frac{\tan \theta }{{\sin }^2\theta }+x$ and $y=x\frac{\tan \theta }{{\sin }^2\theta }-x$

Slope of first line that is $\tan {\theta }_1=\frac{\tan \theta }{{\sin }^2\theta }+1$

Slope of second line $\tan {\theta }_2=\frac{\tan \theta }{{\sin }^2\theta }-1$

$\tan {\theta }_1-\tan {\theta }_2=\frac{\tan \theta }{{\sin }^2\theta }+1-(\frac{\tan \theta }{{\sin }^2\theta }-1)=2$

Hence proved