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My teacher was teaching co-ordinate geometry and today he said that the following equation will always represent a conic section:$$ax^2+by^2+2hxy+2gx+2fy+c=0$$ Then he said that if the determinant of the following matrix is zero then the equation can be split into two linear factors, and thus will represent a couple of straight lines.$$ \begin{bmatrix}a & h & g\\h & b & f\\ g & f & c\end{bmatrix}$$ What is the proof of this theorem?

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  • $\begingroup$ A very useful observation is that if that matrix is $A$ and we set $v = (\begin{matrix}x & y & 1 \end{matrix})^T$, then the conic is given by the equation $$v^T A v = 0$$ $\endgroup$ – Hurkyl Aug 11 '14 at 16:53
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When a conic consists of two lines, moreover called a degenerate conic, it can be written as follows:

$$ (ax+by+c)(dx+ey+f)=0 $$

Its determinant is (MathGem's hint 2):

$$ \frac{1}{8}\begin{vmatrix} 2ad & ae+bd & af+cd \\ ae+bd & 2be & bf+ce \\ af+cd & bf+ce & 2cf\end{vmatrix} $$

which is $0$.

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Hint:

  1. Ask yourself when a conic section becomes degenerate?

  2. Calculate the determinant of that matrix.

Can you see a relation between 1 and 2?

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