UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 10

Problem:

Solution:

Wow! What a long question! Okay, warning, long solution ahead.

For part (a), we have done with the trick a lot of times, just use the line through $ (a, 0) $. The line is $ y = t(a - x) $ (for algebraic simplicity I pick this instead of $ y = t(x - a) $). Putting this back to the equation we get


$ \begin{eqnarray*} y^2(a + x) &=& (a - x)^3 \\ (t(a - x))^2(a + x) &=& (a - x)^3 \\ t^2(a + x) &=& (a - x) \\ (1 + t^2)x &=& a(1 - t^2) \\ x &=& a\frac{1 - t^2}{1 + t^2} \\ y &=& t(a - x) \\ &=& t(a - a\frac{1 - t^2}{1 + t^2}) \\ &=& ta(1 - \frac{1 - t^2}{1 + t^2}) \\ &=& ta(\frac{1 + t^2}{1 + t^2} - \frac{1 - t^2}{1 + t^2}) \\ &=& ta(\frac{2t^2}{1 + t^2}) \end{eqnarray*} $

It looks like t - substitution again? Indeed, again, there is a relation, see my another earlier post to find that out!

For part (b), the key is to use congruent triangles as in the diagram shown

It is now obvious to see that $ \triangle JLP \cong \triangle QGB $, as we already know the x coordinate of $ Q $ is simply $ x $, the key is to find the y coordinate of $ Q $ and this is same as the length as $ JL $.

To find $ JL $, we use $ \triangle JCB \sim \triangle PEB $, we know $ \frac{|PE|}{|EB|} = \frac{|JC|}{|CB|} $, so we have

$ \begin{eqnarray*} |JL| &=& |JC| - |CL| \\ &=& |JC| - |PE| \\ &=& \frac{|PE| \times |CB|}{|EB|} - |PE| \\ &=& \frac{\sqrt{a^2 - x^2} \times 2a}{a + x} - \sqrt{a^2 - x^2} \\ &=& \frac{\sqrt{a^2 - x^2} \times 2a}{a + x} - \frac{(a + x)\sqrt{a^2 - x^2}}{a+x} \\ &=& \frac{\sqrt{a^2 - x^2} \times (a - x)}{a + x} \\ \end{eqnarray*} $

Now we have found the coordinate $ Q $, next we simply substitute that into the equation and prove that it is indeed on the cissoid.
$ \begin{eqnarray*} & & y^2(a + x) \\ &=& \left(\frac{\sqrt{a^2 - x^2} \times (a - x)}{a + x}\right)^2(a + x) \\ &=& \frac{(a^2 - x^2) \times (a - x)^2}{(a + x)^2}(a + x) \\ &=& \frac{(a + x)(a - x) \times (a - x)^2}{(a + x)^2}(a + x) \\ &=& (a - x)^3 \\ \end{eqnarray*} $

So the point $ Q $ indeed lie on the cissoid.

For part (c), we use two point form to compute the prescribed line as $ \frac{y - 0}{x - a} = \frac{\frac{a}{2} - 0}{0 + a} $ which simplifies to $ x = 2y - a $.

Now we substitute this into the first x of the cissoid formula, we get

$ y^2(2y - a + a) = (x - a)^3 $ which readily simplify to $ 2 = \left(\frac{x - a}{y}\right)^3 $.

Phew, what a long question!