# Thread: Parametric equation / Cartesian equation

1. ## Parametric equation / Cartesian equation

My textbook says:

x = 2 + 5 tan t
y = 1 + 3 sec t

By performing appropriate operations on the parametric equations, eliminate the parameter and get a Cartesian equation relating x and y. You should find that one of the Pythagorean properties will help.

I believe the question is asking for me to take the two equations and make one equation in the f(x)=x format so it may be put into a graphing calculator. If this is true, I am 100% at a loss; I cannot find any reference in the chapter that tells how to do this, and as a last resort, I even checked all of the example problems... none relate to this.

2. Take $\sin^{2}(t)+\cos^{2}(t)=1$, and divide the entire equation through by $\cos^{2}(t)$ to get an identity that might help.

3. Originally Posted by wiseguy
My textbook says:

x = 2 + 5 tan t
y = 1 + 3 sec t

By performing appropriate operations on the parametric equations, eliminate the parameter and get a Cartesian equation relating x and y. You should find that one of the Pythagorean properties will help.

I believe the question is asking for me to take the two equations and make one equation in the f(x)=x format so it may be put into a graphing calculator.
Yes, that's what you are asked to do.

If this is true, I am 100% at a loss; I cannot find any reference in the chapter that tells how to do this, and as a last resort, I even checked all of the example problems... none relate to this.
You have the following system of equations:

$x=2+5\frac{\sin t}{\cos t}$ and $y=1+3\frac{1}{\cos t}$

Given the second equation, you can express $\cos t$ in terms of $y$.

You can then replace $\scriptstyle \cos t$ (and even $\scriptstyle \sin t=\pm \sqrt{1-\cos^2 t}$) in the first equation by terms in which $t$ does not occur anymore. Finally, solve that remaining equation (in x and y) for y.

4. Okay, I somewhat overlooked that I may of taken the three variable system of equations and solved for t, then y.

x=2+5tan(t)
(x-2)/5=tanT
arctan((x-2)/5)=T

y=1+3secT
1/3=1/cosT
3/1=cosT
arccos3=T

arctan((x-2)/5)=arccos3
(arctan((x-2)/5)/(arccos3)=y

Am I headed in the right direction with this work?

5. (Ignore this, sorry. Something went wrong in my work.)

6. I think you might be able to get that approach to work. But if you do what I suggested, that is, see that $\tan^{2}(t)+1=\sec^{2}(t),$ then you simply solve each of your two equations for the relevant trig function, and drop them into the equation I just wrote down. Voila, you're done.

7. x=2+5sinT/cosT
(x-2)/5sinT=1/cosT;
y=1+3(1/cosT)
(y-1)/3=1/cosT;
(x-2)/5sinT=(y-1)/3

3(x-2)/5sinT+1=y

I don't know if I screwed that up or not...

I'm a little confused about using the tan^2 identity :/ would be nice to impress the teacher, lol.

8. What Ackbeet was suggesting was what I erased -- I ended up getting two possible equations when I tried to solve for y, and I thought I was missing something. I'll just show you the beginning part:

Solve each equation for the corresponding trig function:
\begin{aligned}
tan\, t &= \frac{x - 2}{5} \\
sec\, t &= \frac{y - 1}{3}
\end{aligned}

Use the pythagorean identity $1 + tan^2\, t = sec^2\, t$ that Ackbeet mentioned and plug in :
\begin{aligned}
1 + tan^2\, t &= sec^2\, t \\
1 + \left( \frac{x - 2}{5} \right)^2 &= \left( \frac{y - 1}{3} \right)^2 \\
\end{aligned}

EDIT: Duh, this is an equation for a hyperbola. Who says we need to solve for y? I'm such an idiot. So just put it in standard form:
\begin{aligned}
1 + \left( \frac{x - 2}{5} \right)^2 &= \left( \frac{y - 1}{3} \right)^2 \\
\frac{(y - 1)^2}{9} - \frac{(x - 2)^2}{25} &= 1
\end{aligned}

9. Hello, wiseguy!

$\begin{array}{cccc}x &=& 2 + 5 \tan t & [1]\\
y &=& 1 + 3\sec t & [2] \end{array}$

Eliminate the parameter and get a Cartesian equation in $x$ and $y$.

$\begin{array}{cccccc}\text{Rewrite [2]:} & \dfrac{y-1}{3} &=& \sec t & [3] \\ \\[-3mm] \text{Rewrite [1]:} & \dfrac{x-2}{5} &=& \tan t & [4] \end{array}$

. . $\begin{array}{cccccc}
\text{Square [3]:} & \left(\dfrac{y-1}{3}\right)^2 &=& \sec^2t & [5] \\ \\[-3mm]
\text{Square [4]:} & \left(\dfrac{x-2}{5}\right)^2 &=& \tan^2t & [6] \end{array}$

$\text{Subtract [5] - [6]: }\;\left(\dfrac{y-1}{3}\right)^2 - \left(\dfrac{x-2}{5}\right)^2 \;=\;\underbrace{\sec^2t - \tan^2t}_{\text{This is 1}}$

$\text{Therefore: }\;\dfrac{(y-1)^2}{9} - \dfrac{(x-2)^2}{25} \;=\;1$

10. Okay, so it doesn't matter that the equation needs to be solved for Y, $(x-h)^2/a^2+(y-k)2 /b^2 = 1$ works. Maybe I'll be the only one in the class who figured this tidbit out and I'll impress the instructor