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 $.
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 $.
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