UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 8

Problem:

Solution:

Part (a) is simple. First of all, notice we have a $ y^2 $ on the left hand side, so normal polynomial reasoning does not apply. The easiest way to do this problem is geometrically.

The line x = d > c cuts the curve at 0 points
The line x = 0 cuts the curve at one point.
The line x = c > e  > 0 cuts the curve at two points.
The line y = 0.1 (assuming it is still the loop area, depending of course on c) cuts the curve at 3 points.

Part (b) is also simple, I will skip the picture. The line through the origin has formula $ y = mx $, so if we substitute this into the formula, we get

$ (mx)^2 = cx^2 - x^3 $.

Then we divide $ x^2 $ throughout to get $ m^2 = c - x $, that is why we get exactly one other point on the curve because the equation will have exactly one solution.

For part (c), just replace $ m $ by $ t $ above. If $ t^2 = c $, we will get the origin. If $ t^2 < c $, we will get positive $ x $ and therefore the loop region, otherwise we will get negative $ x $ and therefore the tails, so we get the whole curve.

This is somewhat unusual, normally parametrization miss some points, but for this one we covered all, perhaps because we do not have a denominator in this case.

For Part (d), we just need to do a little more algebra, we already know $ t^2 = c - x $, so $ x = c - t^2 $, $ y = tx = t(c - t^2) $, that's it!