Why is $h^2-ab$ the discriminant of the conic with equation $ax^2+2hxy+by^2+2gx+2fy+c=0$?

I recently began my study of conic sections at high school. the term $$h^2-ab$$ was declared the discriminant of equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ (which represents a conic or pair of lines). I tried my best to derive the given discriminant from the given standard equation of any conic but all in vain. A good reasoning behind why $$h^2-ab$$ is the discriminant of the standard equation of any conic would be appreciated.

• Are you familiar with the focus-directrix definition of a conic? – Blue Apr 30 '20 at 18:25
• May help: brilliant.org/wiki/conics-discriminant However, it's unclear what you don't understand. – Jean-Claude Arbaut Apr 30 '20 at 18:27
• Let’s start with this: do you understand what the value of the discriminant tells you about the conic? – amd Apr 30 '20 at 19:47
• If you pretend either $x$ or $y$ is a constant then the $ax^2+2hxy+by^2$ is just a quadratic equation. – CyclotomicField May 1 '20 at 4:51
• I am aware of the focus-directrix definition of a conic sir. @Blue. – Huraira majeed May 1 '20 at 20:05

The following development is based on the given equation represents a pair of straight line only.

By suitable translation, the given pair of straight lines is pair-wisely parallel to $$ax^2 + 2hxy + by^2= 0$$ Equivalently, we have a reduced equation:

$$(\dfrac yx)^2 + \dfrac {2h}{b}(\dfrac {y}{x}) + \dfrac {a}{b} = 0$$

If it represents a pair of straight lines, then we should have a pair of straight line passing through the origin and hence $$(\dfrac {y}{x} – m_1)( \dfrac {y}{x} – m_2) = 0$$ where $$m_1$$ and $$m_2$$ are the slopes of those two lines.

On the other hand, $$m_1$$, and $$m_2$$ are the roots of the reduced equation. Then, $$m_1 + m_2 = ….$$ and $$m_1m_2 = ….$$.

Note that if $$\theta$$ is the angle between these two lines, then

$$\tan \theta = (+/-) \dfrac {m_1 – m_2}{1 + m_1m_2} = … = (+/-) \dfrac {2\sqrt {h^2 – ab}}{a + b}$$

The values a, b and h can affect $$h^2 – ab$$. That, in turn, can be used to determine $$\theta$$ and hence the condition of those two lines.

• thanks a lot sir,@Mick – Huraira majeed May 2 '20 at 18:29