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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Are you familiar with the focus-directrix definition of a conic? – Blue Apr 30 '20 at 18:25
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2May help: https://brilliant.org/wiki/conics-discriminant/ However, it's unclear what you don't understand. – Jean-Claude Arbaut Apr 30 '20 at 18:27
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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
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If you pretend either $x$ or $y$ is a constant then the $ax^2+2hxy+by^2$ is just a quadratic equation. – CyclotomicField May 01 '20 at 04:51
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I am aware of the focus-directrix definition of a conic sir. @Blue. – Huraira majeed May 01 '20 at 20:05
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I know sir that value of discriminant tells us the type of conic @amd – Huraira majeed May 01 '20 at 20:08
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@Hurairamajeed: Okay, then consider a conic with focus $(p,q)$, directrix $x\cos\theta+y\sin\theta=r$, and eccentricity $\varepsilon$. Determine the general second-degree equation for this conic and see what the corresponding value for $h^2-ab$ tells you. – Blue May 01 '20 at 20:08
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@Blue, h^2-ab equals e^2-1......... i dont know what this expression means sir. – Huraira majeed May 01 '20 at 20:44
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@Hurairamajeed: Very good! Now, all you need to do is observe that the sign of $h^2-ab$ directly reflects the relative size of $\varepsilon$ and $1$, which in turn characterizes the type of conic involved. For instance, $$h^2-ab < 0 \qquad\Leftrightarrow\qquad \varepsilon < 1 \qquad\Leftrightarrow\qquad \text{ellipse}$$ – Blue May 01 '20 at 22:02
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Hi @Blue, sir can you explain from the equation $ax^2+by^2+2hxy+2gx+2fy+c=0$, How does sign of $h^2-ab$ decide the conic? For example, if I consider it as a quadratic in $x$, then it would become $$x=\dfrac{-(hy+g)}{a} \pm \dfrac{1}{a}\sqrt{(h^2-ab)y^2+2(gh-af)y+g^2-ac}$$. Now in the square root, it's a quadratic equation, so the sign of $h^2-ab$ would decide if it would be an upward parabola, or downward parabola, and it will be $0$ also. I cannot analyse how different signs yield different conics. – V.G Mar 18 '21 at 11:50
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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.
Mick
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I want to know why it's true for all two degree curves, that is, why is it a parabola if $h^2=ab$, ellipse if $h^2<ab$, and hyperbola if $h^2>ab$ given that discriminant is positive. – V.G Mar 18 '21 at 11:42
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@LightYagami This post is discussing for the case of a pair of atraight lines only. For yur question on conditions that a general second degree equation to represent which conic, I suggest you should see this post:- https://math.stackexchange.com/questions/1769353/how-to-determine-standard-equation-of-a-conic-from-the-general-second-degree-equ – Mick Mar 19 '21 at 09:06