Why does the angle between a Pair of Straight Lines depend only on the Homogeneous Part? The most general form of a quadratic is: $ax^2 + by^2 + 2gx + 2fy + c + 2hxy= 0 $ and that for a homogeneous second degree equation is : $ax^2 + by^2 + 2hxy=0$ 
I derived the formula $$\tan \theta = \left|\dfrac{2\sqrt{h^2 - ab}}{a+b}\right|$$ 
but later, author claimed that the same formula is applicable for the general form too. 
But I (and he too) derived it for the homogeneous case only. How can he claim this then?
 A: One can write $$0 = ax^2+by^2+2gx+2fy+c+2hxy = (a_1x+b_1y+c_1)(a_2x+b_2y+c_1)$$
So, the angle between two lines is $$\tan \theta = \left|\frac{\tan \alpha -\tan \beta}{1+\tan\alpha\tan\beta}\right| = \left|\frac{-\frac{a_1}{b_1}+\frac{a_2}{b_2}}{1+\frac{a_1a_2}{b_1b_2}}\right| = \left|\frac{-a_1b_2+b_1a_2}{a_1a_2+b_1b_2}\right|= \left|\frac{\sqrt{(a_1b_2+a_2b_1)^2-4a_1a_2b_1b_2}}{a+b}\right| = \left|\frac{\sqrt{4h^2-4ab}}{a+b}\right|.$$
A: On varying the constant terms, we actually translate the line parallel to itself, or in other words the slope is not dependant on the constant terms. Angle between two set of parallel lines is same. 
Just play with the following graph, you will definitely understand. 
Click here to open the graph in Desmos.
Hope you played with the graph. Now, the angle between pair of straight lines does not depend upon the value of the constant terms. Since, we have already found the angle between pair of straight lines represented by the homogeneous equation. It is reasonable to claim, this is the angle for any pair of straight lines represented by general equation $ax^2+2hxy+by^2+2gx+2fy+c=0.$
Moral of the story:

The angle between a pair of straight lines represented by $ax^2+2hxy+by^2+2gx+2fy+c=0$ depends only on the homogeneous part $ax^2+2hxy+by^2=0$.

A: The slopes of the lines are unaffected by the linear and constant terms of the equation. To see this, find the intersection point of the lines and translate the origin to this point. Doing so will eliminate the linear terms of the equation. This intersection can be found in various ways†, such as setting the partial derivatives to zero, yielding $$x_0={bg-fh\over h^2-ab}, y_0={af-gh\over h^2-ab}.$$ Setting $x=x'+x_0$ and $y=y'+y_0$ in the general equation produces $$ax'^2+by'^2+2hx'y'+{abc-af^2-bg^2+2fgh-ch^2\over h^2-ab}=0.$$ The numerator of the constant term is equal to the determinant of $$\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}$$ which vanishes since this is a degenerate conic, leaving $$ax'^2+by'^2+2hx'y'=0.$$

† Another way is to perform Gaussian elimination on the above coefficient matrix. This will both give you the center point and also show that the constant term in the transformed equation must vanish for this null space to be non-trivial.
A: For $2^\circ$ homogeneous equation $ax^2+2hxy+by^2=0$, we say $m_1+m_2=-\dfrac{2h}{b}$ where $m_1$ is the slope of first line and $m_2$ is the slope of second line.
Proof:
$$ax^2+2hxy+by^2=0$$
$$a+2h\dfrac{y}{x}+b\dfrac{y^2}{x^2}=0$$
$$bm^2+2hm+a=0$$
$$m_1+m_2=-\dfrac{2h}{b}$$
But if we talk about general $2^\circ$ equation $ax^2+2hxy+by^2+2gx+2fy+c=0$, then also we can say $m_1+m_2=-\dfrac{2h}{b}$
Here is the proof:
$ax^2+2hxy+by^2+2gx+2fy+c=0$ can be written as 
$$\left(x+\dfrac{hy-y\sqrt{h^2-ab}}{a}+c_1\right)\left(x+\dfrac{hy+y\sqrt{h^2-ab}}{a}+c_2\right)=0$$
Slope of first line is 
$$m_1=\dfrac{-a}{h-\sqrt{h^2-ab}}=-\dfrac{h+\sqrt{h^2+ab}}{b}$$
$$m_2=\dfrac{-a}{h+\sqrt{h^2-ab}}=-\dfrac{h-\sqrt{h^2+ab}}{b}$$
$$m_1+m_2=-\dfrac{2h}{b}$$
