UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 1

Problem:

Solution:

(Part a)

It is an ellipse, let me put it in a standard form.

$ \begin{eqnarray*} x^2 + 4y^2 + 2x - 16y + 1 &=& 0 \\ x^2 + 2x + 4y^2 - 16y &=& -1 \\ x^2 + 2x + 1 + 4y^2 - 16y + 16 &=& -1 + 1 + 16 \\ x^2 + 2x + 1 + 4(y^2 - 4y + 4) &=& 16 \\ (x + 1)^2 + 4(y - 2)^2 &=& 16 \\ \frac{(x + 1)^2}{4^2} + \frac{(y - 2)^2}{2^2} &=& 1 \\ \end{eqnarray*} $.

Therefore it is an ellipse with center $ (-1, 2) $ and radii on x-axis being 4, radii on y-axis being 2, it looks like this:

It can be parameterized with one parameter, so it should be 1 dimensional.

As an aside, if I add a 'curve' on this plot on excel, it looks like an lemon :(

(Part b)

This is really just $ x = \pm y $, therefore, it is the pair of straight line $ x = y $ and $ x = -y $.

I am not sure about the dimension of this one, but look like we cannot do that in one parameter, so at least two.

(Part c)

Let's solve the pair of equations

$ 2x + y - 1 = 0 $
$ 3x - y + 2 = 0 $.

The second equation shows $ y = 3x + 2 $, so just substitute that back to equation 1 to get

$ 2x + 3x + 2  - 1 = 0 $, solving get $x = -\frac{1}{5} $, $ y = \frac{7}{5} $

So this is simply the point $ (\frac{-1}{5}, \frac{7}{5}) $.

Just a point has dimension 0.