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

Question

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

only in the first quadrant. Otherwise it seems like y=-x is an asymptote. — , Aug 14 '19 at 15:53

Amit Rajaraman · Accepted Answer · 2019-08-14T16:00:49.927

8

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.

edited Aug 14 '19 at 16:00

answered Aug 14 '19 at 15:54

Amit Rajaraman

1

It only remains to verify that a vertical line through the origin is not a solution either. – hmakholm left over Monica Aug 14 '19 at 15:57
Edited my answer to include that. – Amit Rajaraman Aug 14 '19 at 16:01

David G. Stork · Answer 2 · 2019-08-14T17:33:16.290

3

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

edited Aug 14 '19 at 17:33

answered Aug 14 '19 at 16:22

David G. Stork

2

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 '19 at 16:31
@MishaLavrov: Sure... but that won't change the answer. – David G. Stork Aug 14 '19 at 16:32
1

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 '19 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 '19 at 16:51
@MishaLavrov : ContourPlot[x^3 + y^3 == 8, ... ] also works. – Eric Towers Aug 14 '19 at 16:53

2 Answers2