Question

If the equation of the lines passing through point $(1,1)$, one making an angle $\theta$ with the positive direction of $x-$axis and the other making the same angle with the positive direction of $y-$axis, is $x^2-(a+2)xy+y^2+a(x+y-1)=0,a\ne -2,$ then the value of sin $2\theta$ is

• A. $a-2$
• B. $a+2$
• C. $\frac{2}{(a+2)}$
• D. $\frac{2}{a}$

Answer 1

The correct option is C $\frac{2}{(a+2)}$

Let $\tan \theta =m$
Equations of the given lines are $y-1=m(x-1)$ and
$y-1=\frac{1}{m}(x-1)$
So their combined equation is
$(y-1{)}^2+(x-1{)}^2-[(x-1)(y-1)](m+\frac{1}{m})=0$
$\Rightarrow x^2+y^2-2(x+y-1)-(m+\frac{1}{m})xy+(m+\frac{1}{m})(x+y-1)=0\Rightarrow x^2+y^2-(m+\frac{1}{m})xy+(m+\frac{1}{m}-2)(x+y-1)=0$
Comparing with the given equation, we get
$a+2=m+\frac{1}{m}=\tan \theta +\cot \theta \Rightarrow \frac{1}{\sin \theta \cos \theta }=a+2\Rightarrow \sin 2\theta =\frac{2}{a+2}$