$x^3+y^3=8$ ,Number of straight lines through origin which do not meet this curve is

Sorry to say that I have no approach for this question, my Brain is totally blank right now. Plz help me.

  • $\begingroup$ Make the ansatz $y=mx$ $\endgroup$ – Dr. Sonnhard Graubner Aug 14 at 15:48
  • $\begingroup$ only in the first quadrant. Otherwise it seems like y=-x is an asymptote. $\endgroup$ – Matthew Daly Aug 14 at 15:53

If a line, say, $y=mx$ does not meet the curve, then $x^3 + (mx)^3 = 8$ has no solutions in $x$.

For $m\neq -1$, we can divide by $1+m^3$ to get a solution. For $m=-1$, there is no solution.

The case of a vertical line through the origin, $x=0$ can easily be verified to intersect the curve.

  • 1
    $\begingroup$ It only remains to verify that a vertical line through the origin is not a solution either. $\endgroup$ – Henning Makholm Aug 14 at 15:57
  • $\begingroup$ Edited my answer to include that. $\endgroup$ – Epiksalad Aug 14 at 16:01

All such lines with slope other than -1 meet the curve:

enter image description here

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    $\begingroup$ Use the CubeRoot command in Mathematica (plotting CubeRoot[8 - x^3]) to include the part of the curve with $x>2$ as well. $\endgroup$ – Misha Lavrov Aug 14 at 16:31
  • $\begingroup$ @MishaLavrov: Sure... but that won't change the answer. $\endgroup$ – David G. Stork Aug 14 at 16:32
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    $\begingroup$ Well, right now it looks like a line with slope $-\frac12$ doesn't intersect the curve, even though it does, at $x \approx 2.091$. $\endgroup$ – Misha Lavrov Aug 14 at 16:34
  • $\begingroup$ All lines with slope less than $-1$ meet the part of the curve you have plotted. Only lines with slope $\geq 0$ meet the part of the curve you have plotted. $\endgroup$ – Eric Towers Aug 14 at 16:51
  • $\begingroup$ @MishaLavrov : ContourPlot[x^3 + y^3 == 8, ... ] also works. $\endgroup$ – Eric Towers Aug 14 at 16:53

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