If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2$ $=0, \mathrm{k}>0$ is $45^{\circ}$, then…
If the angle between the pair of lines $x^2+2 \sqrt{2} x y+k y^2$ $=0, \mathrm{k}>0$ is $45^{\circ}$, then the area (in square units) of the triangle formed by the pair of bisectors of the angles between these lines and the line $x+2 y+1=0$ is
$\frac{1}{3}$
1
$\frac{2}{3}$
2
Solution
Given: $x^2+2 \sqrt{2} x y+k y^2=0$ ...(i)
$x+2 y+1=0$ ...(ii)
The angle between the lines (i) is $45^{\circ}$.
$\therefore \cos 45^{\circ}=\left|\frac{1+k}{\sqrt{(1-k)+8}}\right| \Rightarrow \frac{1}{\sqrt{2}}=\left|\frac{1+k}{\sqrt{9-k}}\right|$
$\begin{aligned} & \Rightarrow \frac{1}{2}=\frac{(1+k)^2}{(9-k)} \Rightarrow 2(1+k)^2=(9-k) \\ & \Rightarrow 2\left(1+k^2+2 k\right)=9-k \\ & \Rightarrow 2 k^2+5 k-7=0 \\ & \Rightarrow 2 k^2+7 k-2 k-7=0 \\ & \Rightarrow(2 k+7)(k-1)=0 \Rightarrow k=\frac{-7}{2}, 1\end{aligned}$
But $k>0 \Rightarrow k=1$
The equation of angular bisector of (i) is
$\begin{aligned} & \Rightarrow h\left(x^2-y^2\right)-(a-b) x y=0 \\ & \Rightarrow \quad \sqrt{2}\left(x^2-y^2\right)-(1-k)=0\end{aligned}$ ...(3)
The required area formed by the lines (ii), (iii) is
$\frac{n^2 \sqrt{h^2-a b}}{\left|a m^2-2 h l m+b l^2\right|}=\frac{1}{3}$.