Find the angle between the straight lines joining the origin to the intersections of the line $\;$ $y = 2x + 1$, $\;$ with the curve $\;$ $x^2 - 3xy + 2y^2 -4x -3y+ 1 = 0$
Equation of given curve is: $\;$ $x^2 - 3xy + 2y^2 -4x -3y+ 1 = 0$ $\;\;\; \cdots \; (1)$
Equation of given curve is: $\;$ $y = 2x + 1$ $\;\;\; \cdots \; (2)$
Equation $(2)$ can be written as
$y - 2x = 1$ $\;\;\; \cdots \; (2a)$
$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,
$x^2 - 3xy + 2y^2 - 4 \left(y - 2x\right)x - 3 \left(y - 2x\right)y + 1 \times \left(y - 2x\right)^2 = 0$
i.e. $\;$ $x^2 -3xy + 2y^2 - 4xy + 8x^2 - 3y^2 + 6xy + y^2 + 4x^2 - 4xy = 0$
i.e. $\;$ $13x^2 - 5xy + 0y^2 = 0$ $\;\;\; \cdots \; (3)$
Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.
Comparing equation $(3)$ with the standard equation $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives
$a = 13, \; h = \dfrac{-5}{2}, \; b = 0$
Let $\theta$ be the angle between the lines given by equation $(3)$.
Then, $\;$ $\tan \theta = \dfrac{2 \sqrt{h^2 - ab}}{a + b}$
Here, since $b = 0$, $\;$ $\tan \theta = \dfrac{2h}{a}$
i.e. $\;$ $\tan \theta = \dfrac{2 \times \left(\dfrac{-5}{2}\right)}{13} = \dfrac{-5}{13}$
$\implies$ $\theta = \tan^{-1} \left(\dfrac{-5}{13}\right)$