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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.

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

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

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