[SOLVED] Parametric Representation of a Curve

**Solo** · May 20th 2009, 02:46 PM

Hey,

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

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

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

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

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 $\frac{-\pi}{2} < t < \frac{\pi}{2}$ ?

Any help would be greatly appreciated!!

**Danny** · May 20th 2009, 03:11 PM

Originally Posted by Solo

Hey,

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

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

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

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

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 $\frac{-\pi}{2} < t < \frac{\pi}{2}$ ?

Any help would be greatly appreciated!!

If you do this, $t= cos^{-1} \frac{1}{x}$ , you're going to get into trouble real fast.

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

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

**Spec** · May 20th 2009, 03:18 PM

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

**Solo** · May 20th 2009, 03:24 PM

Ah yes!

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

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

And the curve is a hyperbola?

As a side note, I have a few questions:

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

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

What implications does that have on the curve when drawn?

**Solo** · May 20th 2009, 03:31 PM

Originally Posted by Spec

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

Hmm I see this gives the same result but I don't see how this was deduced: $y=\tan\left(\arccos\frac{1}{x}\right)=\frac{\sqrt{ 1-\frac{1}{x^2}}}{\frac{1}{x}}=$ $\sqrt{x^2-1}$

**Spec** · May 20th 2009, 03:42 PM

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.

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