UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 5

Problem:

Solution:

For part (a), just substitute the definition of the hyperbolic trigonometric function into the hyperbola.

$\begin{eqnarray*} & & x^2 - y^2 \\ &=& \cosh^2(t) - \sinh^2(t) \\ &=& (\frac{1}{2}(e^t + e^{-t}))^2 - (\frac{1}{2}(e^t - e^{-t}))^2 \\ &=& \frac{1}{4}((e^{2t} + 2 + e^{-2t}) - (e^{2t} - 2 + e^{-2t})) \\ &=& \frac{1}{4}(4) \\ &=& 1 \end{eqnarray*}$

Note that the hyperbola cover the whole y range, we also know that hyperbolic sine is monotonic and cover the whole $y$ range, so we know for each $y$ value, there is a corresponding $t$, so we have a corresponding $x$ as well. Note that as hyperbolic cosine is always positive, so it never covers the whole left hand side of the hyperbola.

For part (b), $x = 0$ does not touch the hyperbola, $x = 1$ touches the hyperbola once as a tangent, $y = 0$ cuts the hyperbola twice.

For part (c), consider the line $y = t(x + 1) = tx + t$, substitute this into the hyperbola, we get:

$\begin{eqnarray*} x^2 - y^2 &=& 1 \\ x^2 - (tx + t)^2 &=& 1 \\ x^2 - (t^2x^2 + 2t^2x + t^2)&=& 1 \\ (1 - t^2)x^2 - 2t^2x - t^2 &=& 1 \\ (1 - t^2)x^2 - 2t^2x - (t^2 + 1) &=& 0 \end{eqnarray*}$

Normally, we would have to solve this, but we already know one of the solution must be $(-1, 0)$, so we can simply use the sum of root formula.

$\begin{eqnarray*} -1 + x &=& \frac{-b}{a} = \frac{2t^2}{1-t^2} \\ x &=& \frac{2t^2}{1-t^2} + 1 \\ &=& = \frac{1 + t^2}{1 - t^2} \end{eqnarray*}$

Then we simply also computes $y$.

$\begin{eqnarray*} y &=& tx + t = t\frac{1 + t^2}{1 - t^2} + t \\ &=& \frac{t + t^3}{1 - t^2} + \frac{t-t^3}{1 - t^2} \\ &=& \frac{2t}{1 - t^2} \end{eqnarray*}$

The final parametrization is therefore $\left(\begin{array}{c}x \\ y\end{array}\right) = \left(\begin{array}{c}\frac{1 + t^2}{1 - t^2} \\ \frac{2t}{1 - t^2}\end{array}\right)$. It is interesting to compare it with the circle's parametrization as $\left(\begin{array}{c}x \\ y\end{array}\right) = \left(\begin{array}{c}\frac{1 - t^2}{1 + t^2} \\ \frac{2t}{1 + t^2}\end{array}\right)$

Part (d) is obvious now, of course the line cannot be the asymptote, that correspond to $t = \pm 1$.