trigonometry - Prove that the envelope of the family of lines $(costheta+sintheta)x+(costheta-sintheta)y+2sintheta-costheta-4=0$

Prove that the envelope of the family of lines $(\cos\theta+\sin\theta)x+(\cos\theta-\sin\theta)y+2\sin\theta-\cos\theta-4=0$

I did not know much about how to find envelope of a curve.I read on Wolfram and tried solving but did not get the desired answer.

I partially differentiated $(\cos\theta+\sin\theta)x+(\cos\theta-\sin\theta)y+2\sin\theta-\cos\theta-4=0$ wrt $\theta$,getting

$(\cos\theta-\sin\theta)x-(\cos\theta+\sin\theta)y+2\cos\theta+\sin\theta=0$

then i squared and added them but could not eliminate $\theta$ fully.Is my method correct?

Please help me.

Answer

HINT

I would say, equation and its derivative together add up and subtract $\Rightarrow$ after simplification two equations:

$(2 x+1) \sin(\theta)+(2 y-3) \cos(\theta)=4$ , $\quad (2 x+1) \cos(\theta)-(2 y-3) \sin(\theta)=4$

I am sure that you can take from here.