Find the asymptotes of the curve $(2x-3y+1)^2(x+y)-8x+2y-9=0$ and show that they intersect the curve again in three points which lie on a straight line.

The asymptotes I found are $x+y=0; 2x-3y=1; 2x-3y=-3$ Except for the first one, none are exactly intersecting the curve. Where am I going wrong?

  • $\begingroup$ To me, you are perfectly correct. $\endgroup$ – Claude Leibovici Jul 26 '17 at 5:03
  • $\begingroup$ Thanks! But the first asymptote i.e x+y=0 gives the point of intersection with curve: (-9/10, 9/10). While the other two asymptotes i.e 2x-3y=1 & 2x-3y=-3 do not give any points of intersection with curve. The question however asks for 3 points of intersection of asymptotes with the curve that lie on a straight line. How do I get the other two points? $\endgroup$ – user464068 Jul 26 '17 at 5:45

The line they are talking about is in a projective sense. The other two point of intersection are at infinite. So the line you are looking for is the line passing through $\left(-\frac{9}{10},\; \frac{9}{10}\right)$ with slope $m=\dfrac{2}{3}$

That is $y-\dfrac{9}{10}=\dfrac{2}{3} \left(x+\dfrac{9}{10}\right)$ or $4 x-6 y+9=0$

As I did it I attach a graph. The violet line is the line passing through the 3 points of intersection. The other are the asymptotes

Hope this helps

enter image description here

  • $\begingroup$ Thanks a lot! But how did you get the slope m? $\endgroup$ – user464068 Jul 26 '17 at 12:16
  • 1
    $\begingroup$ If the lines meet at a infinity point they have the same direction, i.e. the same slope $\endgroup$ – Raffaele Jul 26 '17 at 12:38

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