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If $\theta$ is the acute angle between the lines joining the origin to the points of intersection of the curve $x^2+x y+y^2+x$ $+3 y+1=0$ and the straight line $x+y+2=0$ then $\cos \theta=$
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{5}}$
$x+y+2=0$
$$
\begin{aligned}
& \Rightarrow\left(\frac{x+y}{-2}\right)=1 \Rightarrow x^2+x y+y^2+x \cdot 1+3 y \cdot 1+1^2=0 \\
& \Rightarrow x^2+x y+y^2+x\left(\frac{x+y}{-2}\right)+3 y\left(\frac{x+y}{-2}\right)+\left(\frac{x+y}{-2}\right)^2=0 \\
& \Rightarrow x^2+y^2+x y-\frac{\left(x^2+x y\right)}{2}-\frac{\left(3 x y+3 y^2\right)}{2}+\frac{(x+y)^2}{4}=0 \\
& \Rightarrow 4 x^2+4 y^2+4 x y-2 x^2-2 x y-6 x y-6 y^2+x^2+y^2+2 x y=0 \\
& \Rightarrow 3 x^2-y^2-2 x y=0 \\
& \Rightarrow 3-\left(\frac{y}{x}\right)^2-2\left(\frac{y}{x}\right)=0
\end{aligned}
$$
For $a x^2+2 h x y+b y^2=0$
$$
\begin{aligned}
& \text { Angle between lines } \Rightarrow \tan \theta=\frac{2 \sqrt{h^2-a b}}{|a+b|} \\
& \Rightarrow \tan \theta=\frac{2 \sqrt{1^2-3(-1)}}{|3-1|}=\frac{4}{2} \\
& \tan \theta=2 \Rightarrow \cos \theta=\frac{1}{\sqrt{5}}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow\left(\frac{x+y}{-2}\right)=1 \Rightarrow x^2+x y+y^2+x \cdot 1+3 y \cdot 1+1^2=0 \\
& \Rightarrow x^2+x y+y^2+x\left(\frac{x+y}{-2}\right)+3 y\left(\frac{x+y}{-2}\right)+\left(\frac{x+y}{-2}\right)^2=0 \\
& \Rightarrow x^2+y^2+x y-\frac{\left(x^2+x y\right)}{2}-\frac{\left(3 x y+3 y^2\right)}{2}+\frac{(x+y)^2}{4}=0 \\
& \Rightarrow 4 x^2+4 y^2+4 x y-2 x^2-2 x y-6 x y-6 y^2+x^2+y^2+2 x y=0 \\
& \Rightarrow 3 x^2-y^2-2 x y=0 \\
& \Rightarrow 3-\left(\frac{y}{x}\right)^2-2\left(\frac{y}{x}\right)=0
\end{aligned}
$$
For $a x^2+2 h x y+b y^2=0$
$$
\begin{aligned}
& \text { Angle between lines } \Rightarrow \tan \theta=\frac{2 \sqrt{h^2-a b}}{|a+b|} \\
& \Rightarrow \tan \theta=\frac{2 \sqrt{1^2-3(-1)}}{|3-1|}=\frac{4}{2} \\
& \tan \theta=2 \Rightarrow \cos \theta=\frac{1}{\sqrt{5}}
\end{aligned}
$$
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