Problem:
Solution:
The key identity is $ 1 + \tan^2(t) = \sec^2 t = \frac{1}{\cos^2 t } = \frac{1}{1 - \sin^2 t} $.
With that, we know $ x = a \sin t $ so that $ \sin t = \frac{x}{a} $, put it into the $ y $ equation to get
$ \begin{eqnarray*} y^2 &=& a^2 \tan^2 t (1 + \sin t)^2 \\ &=& a^2 (\sec^2 t - 1) (1 + \sin t)^2 \\ &=& a^2 (\frac{1}{1 - \sin^2 t} - 1) (1 + \sin t)^2 \\ &=& a^2 (\frac{1}{1 - (\frac{x}{a})^2} - 1) (1 + \frac{x}{a})^2 \\ &=& (\frac{1}{1 - (\frac{x}{a})^2} - 1) (a + x)^2 \\ &=& (\frac{a^2}{a^2 - x^2} - 1) (a + x)^2 \\ &=& (\frac{a^2}{a^2 - x^2} - \frac{a^2 - x^2}{a^2 - x^2}) (a + x)^2 \\ &=& \frac{x^2}{a^2 - x^2} (a + x)^2 \end{eqnarray*} $
As the hint specify, we cannot multiply over the denominator because then we will introduce an additional straight line $ x = -a $ in the graph. This seems to illustrate a different point. While we miss points when we find rational parametization to a curve, we introduce new points when we force ourselves to represent this as polynomial (or variety), I still do not know if it is possible to represent this curve as a variety instead of a rational function.
For part (b), let's us try the usual trick the consider the point intersected by the line through origin. For easy of manipulation, we start with the sloppy formula:
$ \begin{eqnarray*} y^2(a^2 - x^2) &=& x^2(a + x)^2 \\ (tx)^2(a^2 - x^2) &=& x^2(a + x)^2 \\ t^2(a^2 - x^2) &=& (a + x)^2 \\ a^2t^2 - t^2x^2 &=& a^2 + 2ax + x^2 \\ (1 + t^2)x^2 + 2ax + a^2(1 - t^2) &=& 0 \\ \end{eqnarray*} $
In the simplification, we already get rid of the double root $ x = 0 $, we know the sloppy formula introduced the line $ x = -a $ so that is not the right answer, consider the sum of roots $ s $:
$ \begin{eqnarray*} x &=& s - (-a) \\ &=& \frac{-2a}{1 + t^2} -(-a) \\ &=& a - \frac{2a}{1 + t^2} \\ &=& \frac{a(1+t^2)}{1 + t^2} - \frac{2a}{1 + t^2} \\ &=& a\frac{t^2 - 1}{1 + t^2} \\ y &=& tx \\ &=& at\frac{t^2 - 1}{1 + t^2} \end{eqnarray*} $
The answer feel like t substitution, so maybe there is some relationship there. Apparently, we can use the t-substitution to the defining formula directly, but it is not the same as what we get here, but if we do a phase shift of $ \frac{\pi}{2} $, then we can get there:
$ \begin{eqnarray*} x &=& a \sin(\theta) \\ &=& -a \cos(\theta + \frac{\pi}{2}) \\ &=& -a \frac{1 - t^2}{1 + t^2} \\ &=& a \frac{t^2 - 1}{1 + t^2} \\ y &=& a \tan \theta (1 + \sin \theta) \\ &=& a \frac{1}{\tan(\theta + \frac{\pi}{2})}(1 - \cos (\theta + \frac{\pi}{2})) \\ &=& a \frac{1}{\frac{2t}{1-t^2}}(1 - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(1 - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(\frac{1 + t^2}{1 + t^2} - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(\frac{-2t^2}{1 + t^2}) \\ &=& at\frac{t^2 - 1}{1 + t^2} \end{eqnarray*} $
So we come to full circle!
Solution:
The key identity is $ 1 + \tan^2(t) = \sec^2 t = \frac{1}{\cos^2 t } = \frac{1}{1 - \sin^2 t} $.
With that, we know $ x = a \sin t $ so that $ \sin t = \frac{x}{a} $, put it into the $ y $ equation to get
$ \begin{eqnarray*} y^2 &=& a^2 \tan^2 t (1 + \sin t)^2 \\ &=& a^2 (\sec^2 t - 1) (1 + \sin t)^2 \\ &=& a^2 (\frac{1}{1 - \sin^2 t} - 1) (1 + \sin t)^2 \\ &=& a^2 (\frac{1}{1 - (\frac{x}{a})^2} - 1) (1 + \frac{x}{a})^2 \\ &=& (\frac{1}{1 - (\frac{x}{a})^2} - 1) (a + x)^2 \\ &=& (\frac{a^2}{a^2 - x^2} - 1) (a + x)^2 \\ &=& (\frac{a^2}{a^2 - x^2} - \frac{a^2 - x^2}{a^2 - x^2}) (a + x)^2 \\ &=& \frac{x^2}{a^2 - x^2} (a + x)^2 \end{eqnarray*} $
As the hint specify, we cannot multiply over the denominator because then we will introduce an additional straight line $ x = -a $ in the graph. This seems to illustrate a different point. While we miss points when we find rational parametization to a curve, we introduce new points when we force ourselves to represent this as polynomial (or variety), I still do not know if it is possible to represent this curve as a variety instead of a rational function.
For part (b), let's us try the usual trick the consider the point intersected by the line through origin. For easy of manipulation, we start with the sloppy formula:
$ \begin{eqnarray*} y^2(a^2 - x^2) &=& x^2(a + x)^2 \\ (tx)^2(a^2 - x^2) &=& x^2(a + x)^2 \\ t^2(a^2 - x^2) &=& (a + x)^2 \\ a^2t^2 - t^2x^2 &=& a^2 + 2ax + x^2 \\ (1 + t^2)x^2 + 2ax + a^2(1 - t^2) &=& 0 \\ \end{eqnarray*} $
In the simplification, we already get rid of the double root $ x = 0 $, we know the sloppy formula introduced the line $ x = -a $ so that is not the right answer, consider the sum of roots $ s $:
$ \begin{eqnarray*} x &=& s - (-a) \\ &=& \frac{-2a}{1 + t^2} -(-a) \\ &=& a - \frac{2a}{1 + t^2} \\ &=& \frac{a(1+t^2)}{1 + t^2} - \frac{2a}{1 + t^2} \\ &=& a\frac{t^2 - 1}{1 + t^2} \\ y &=& tx \\ &=& at\frac{t^2 - 1}{1 + t^2} \end{eqnarray*} $
The answer feel like t substitution, so maybe there is some relationship there. Apparently, we can use the t-substitution to the defining formula directly, but it is not the same as what we get here, but if we do a phase shift of $ \frac{\pi}{2} $, then we can get there:
$ \begin{eqnarray*} x &=& a \sin(\theta) \\ &=& -a \cos(\theta + \frac{\pi}{2}) \\ &=& -a \frac{1 - t^2}{1 + t^2} \\ &=& a \frac{t^2 - 1}{1 + t^2} \\ y &=& a \tan \theta (1 + \sin \theta) \\ &=& a \frac{1}{\tan(\theta + \frac{\pi}{2})}(1 - \cos (\theta + \frac{\pi}{2})) \\ &=& a \frac{1}{\frac{2t}{1-t^2}}(1 - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(1 - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(\frac{1 + t^2}{1 + t^2} - \frac{1 - t^2}{1 + t^2}) \\ &=& a \frac{1-t^2}{2t}(\frac{-2t^2}{1 + t^2}) \\ &=& at\frac{t^2 - 1}{1 + t^2} \end{eqnarray*} $
So we come to full circle!
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