# Thread: [SOLVED] Parametric Representation of a Curve

1. ## [SOLVED] Parametric Representation of a Curve

Hey,

Find the equation in $\displaystyle x$ and $\displaystyle y$ for the curve given parametrically
by $\displaystyle x = sec(t)$ and $\displaystyle y = tan(t)$, where $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$. What type of
curve is it?

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So I've written that $\displaystyle x = \frac{1}{cos(t)}$ and $\displaystyle y = \frac{sin(t)}{cos(t)}$.

From that $\displaystyle t= cos^{-1} \frac{1}{x}$

If I substitute that in $\displaystyle y = \frac{sin (cos^{-1} \frac{1}{x})}{\frac{1}{x}}$.

But that's where I get stuck..I'm not sure if I've even started correctly. What do you with the fact they've told you that $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$?

Any help would be greatly appreciated!!

2. Originally Posted by Solo
Hey,

Find the equation in $\displaystyle x$ and $\displaystyle y$ for the curve given parametrically
by $\displaystyle x = sec(t)$ and $\displaystyle y = tan(t)$, where $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$. What type of
curve is it?

-----
So I've written that $\displaystyle x = \frac{1}{cos(t)}$ and $\displaystyle y = \frac{sin(t)}{cos(t)}$.

From that $\displaystyle t= cos^{-1} \frac{1}{x}$

If I substitute that in $\displaystyle y = \frac{sin (cos^{-1} \frac{1}{x})}{\frac{1}{x}}$.

But that's where I get stuck..I'm not sure if I've even started correctly. What do you with the fact they've told you that $\displaystyle \frac{-\pi}{2} < t < \frac{\pi}{2}$?

Any help would be greatly appreciated!!
If you do this, $\displaystyle t= cos^{-1} \frac{1}{x}$, you're going to get into trouble real fast.

From $\displaystyle x = \frac{1}{cos(t)}$ and $\displaystyle y = \frac{sin(t)}{cos(t)}$, then

$\displaystyle x^2 - y^2 =$ ? Think identities!

3. $\displaystyle \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right.$ $\displaystyle \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$

4. Ah yes!

$\displaystyle x^2 - y^2 = \frac{1}{cos^2(t)} - \frac{sin^2(t)}{cos^2(t)} = \frac{1-sin^2(t)}{cos^2(t)} = \frac{cos^2(t)}{cos^2(t)} = 1$

So Cartesian equation: $\displaystyle x^2 - y^2 = 1$

And the curve is a hyperbola?

As a side note, I have a few questions:

Isn't the equation for a hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$?

So in this case $\displaystyle a = b = 1$?

What implications does that have on the curve when drawn?

5. Originally Posted by Spec
$\displaystyle \left\{\begin{array}{ll}x=\frac{1}{\cos t} \Longleftrightarrow t = \arccos \frac{1}{x}\\y=\tan t \Longleftrightarrow t = \arctan y\end{array}\right.$ $\displaystyle \Longleftrightarrow \arccos \frac{1}{x}=\arctan y \Longleftrightarrow y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$
Hmm I see this gives the same result but I don't see how this was deduced: $\displaystyle y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\displaystyle \sqrt{x^2-1}$

6. List of trigonometric identities - Wikipedia, the free encyclopedia

It's not a correct solution though (one of the equivalence signs are wrong), since we lost all the negative y values.