How does Homogenization of second degree equation to get equation of pair of straight lines work?

Question

Consider this figure:

Here the curve $AB$ is given by $s$, $$s = ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \tag1$$

If equation of line $AB$ is $L$, then, $$L = lx + my + n = 0 \tag2$$

To find the equation of pair of straight lines $OA$ and $OB$, we use the method of homogenization of second degree equation i.e., from $(1)$,

$$s = ax^2 + by^2 + 2hxy + 2gx(1) + 2fy(1) + c(1)^2 = 0 \tag3$$

From $(2)$, $$\frac{lx + my}{-n} = 1 \tag4$$

Now substituting $(4)$ in $(3)$,

$$ax^2 + by^2 + 2hxy + 2gx\left(\frac{lx + my}{-n}\right) + 2fy\left(\frac{lx + my}{-n}\right) + c\left(\frac{lx + my}{-n}\right)^2 = 0 \tag5$$

This way we can homogenize the equation. Homogenizing equation $(1)$ gives us the equation of pair of straight lines $OB$ and $OA$.

My question is how does equation $(5)$ represent the equation of pair of straight lines?

I know that the pair of straight lines passing through the origin are homogenized equation. But how does equation $(5)$ represent that?

---

Is it like:

The equation

• satisfies the point of intersection $A$ and $B$
• is homogeneous i.e. it must be a pair of straight lines passing through the origin

Due to the above reasons we can conclude that the equation is of the pair of straight lines $OA$ and $OB$. Is it?

These were the same reasons as provided in another post.

---

After checking that post, another question struck my mind. Isn't the equation of the curve and the pair of straight line the same as we only substitute equation $(4)$ instead of $1$ ? If yes, how can this be the case?

Answer 1

Doing so we get $$ax^2+b\left(-\frac{lx+n}{m}\right)^2+2hx\left(-\frac{lx+n}{m}\right)+2x\left(-\frac{lx+n}{m}\right)+2f\left(-\frac{lx+n}{m}\right)+c=0$$ This equation can be solved for $x$.

Answer 2

Hint.

Considering

$$ a x^2+b x y + c y^2 = (c_1 x+ c_2 y)(c_3 x + c_4 y) $$

identifying coefficients we get

$$ \cases{ a=c_1 c_3\\ b = c_1c_4+c_2c_3\\ c=c_2c_4 } $$

now solving for $c_1,c_2,c_3$ we have

$$ \cases{ c_1 = \frac{b c_4\pm|c_4|\sqrt{b²-4ac}}{2c_4^2}\\ c_2 = \frac{c}{c_4}\\ c_3 = \frac{b c_4\pm|c_4|\sqrt{b^2-4ac}}{2c} } $$

so the condition for a pair of lines is

$$ b^2-4 a c \ge 0 $$

Answer 3

Since in Eq. (5), the sum of powers of $x$ and $y$ is 2, so it is a homogenous equation of degree 2, notice there is no constant added in it. A homogeneous quadratic may represent just point, pair of real straight lines passing through origin or nothing. $$ \mbox{A point:}~~~x^2+y^2=0 \implies x=0, y=0$$ $$\mbox{A pair of stright lines:} ~~~ax^2+y^2+3xy=0 \implies y=m x, m_1,m_2=\frac{-3\pm \sqrt{5}}{2}.$$ $${\mbox nothing :} ~~~x^2+y^2+xy=0.$$

Answer 4

The equation will only be the same if $$\frac{lx + my}{-n} = 1$$ Which is only true if the point $$(x,y)$$ lies on the line. Since we are substituting ith in the curve and for the point of intersection of the curve and the line, $$\frac{lx + my}{-n} = 1$$ and the equation of curve $$s = ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$$ is also satisfied at that point, we are getting the pair of lines passing through the points and the origin.