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

$$x^3+y^3=8$$ ,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.

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

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

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

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