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For the standard conic equation in 2-D plane i.e $$ax^2 + 2hxy + by^2 +2gx +2fy +c =0$$ If $a=b=0$ and $h\neq 0$, then the equation reduces to $$2hxy + 2gx + 2fy + c =0$$

Under what conditions will the above equation represent a pair of straight lines?

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when $c = \frac{2fg}{h}$

Try a few examples

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  • $\begingroup$ @Sameernilkhan you did not finish your sentence. Also, instead of telling me what you were thinking, try a few examples. $\endgroup$ – Will Jagy Nov 23 '19 at 23:44
  • $\begingroup$ I was thinking about conditions like C1) f=0 and c=0 then the equation becomes 2hxy + 2gx = 0 this implies x(hy + g) = 0 i.e. 2 real pair of straight lines. C2) g=0 and c=0 then the equation becomes 2hxy + 2fy = 0 this implies y(hx + f) = 0 i.e. 2 real pair of straight lines. C3) g=0, f=0 and c=0 then the equation becomes 2hxy=0 this implies x=0 and y=0 i.e. 2 real pair of straight lines. Are any more conditions possible? $\endgroup$ – Sameer nilkhan Nov 23 '19 at 23:59
  • $\begingroup$ I will surely try some examples if you explain the logic behind your condition i. e. c=fg/h. $\endgroup$ – Sameer nilkhan Nov 24 '19 at 0:02
  • $\begingroup$ I checked the example 6xy + 4x + 12y + 4 = 0. It represents a hyperbola. $\endgroup$ – Sameer nilkhan Nov 24 '19 at 0:17
  • $\begingroup$ @Sameernilkhan good, I mistakenly left off a factor of $2.$ Let's see, your $h=3$ $g=2$ $f=6$ so $\frac{2fg}{h}= \frac{24}{3} = 8$ $\endgroup$ – Will Jagy Nov 24 '19 at 0:31
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The equation $2hxy + 2gx + 2fy + c =0$ is expected to be factorized as

$$ (px+q)(ty+s)=0$$

in order for it to present two lines. Match the coefficients to get,

$$pt = 2h, \>\>\> qt = 2g,\>\>\> sp=2f, \>\>\> qs = c$$

Then,

$$\frac pq =\frac hg = \frac {2f}c$$

Thus, the condition is $hc = 2fg$.

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  • $\begingroup$ Thanks i got it. $\endgroup$ – Sameer nilkhan Nov 24 '19 at 1:13

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