# Thread: more parametric equations part 2

1. ## more parametric equations part 2

Eliminate the parameter, t to obtain the cartesian equation in x and y. Sketch the curve, taking note of any restrictions

d) $x = t + \frac{1}{t}, ~ y = t^2 - \frac{1}{t^2}$

e) $x = t + \frac{1}{t}, ~ y= t - \frac{1}{t}$

f) $x = \frac{1 - t^2}{1 + t^2}, ~ y = \frac{2t}{1+t^2}$

I know it's a lot and i really dont expect this all to be answered, i just put all the questions which i had serious trouble with and i would be SO grateful to anyone who could give me a few pointers on how to solve these because i just dont know how to get rid of the t OR find the restrictions. again , i know there are alot of questions and im not just expecting answers or a cheap way to get answers for hw, i realli just dont understand it and would seriously appreciate any help .
thankyou guyz !

2. Originally Posted by THSKluv
Eliminate the parameter, t to obtain the cartesian equation in x and y. Sketch the curve, taking note of any restrictions

d) $x = t + \frac{1}{t}, ~ y = t^2 - \frac{1}{t^2}$

e) $x = t + \frac{1}{t}, ~ y= t - \frac{1}{t}$

f) $x = \frac{1 - t^2}{1 + t^2}, ~ y = \frac{2t}{1+t^2}$

I know it's a lot and i really dont expect this all to be answered, i just put all the questions which i had serious trouble with and i would be SO grateful to anyone who could give me a few pointers on how to solve these because i just dont know how to get rid of the t OR find the restrictions. again , i know there are alot of questions and im not just expecting answers or a cheap way to get answers for hw, i realli just dont understand it and would seriously appreciate any help .
thankyou guyz !
d) Same technique as c) in this thread: http://www.mathhelpforum.com/math-he...-part-1-a.html

e) $x = t + \frac{1}{t} \Rightarrow x^2 = t^2 + 2 + \frac{1}{t^2} \Rightarrow x^2 - 2 = t^2 + \frac{1}{t^2}$ .... (1)

$y= t - \frac{1}{t} \Rightarrow y^2 + 2 = t^2 + \frac{1}{t^2}$ .... (2)

Equate equations (1) and (2).

f) Square each equation. Add. Simplify. This question is similar to e) in this thread: http://www.mathhelpforum.com/math-he...ions-help.html

3. Hello, THSKluv!

These are tricky, requiring special techniques.

Eliminate the parameter $t$ to obtain the cartesian equation in $x$ and $y.$
Sketch the curve, taking note of any restrictions.

$d)\;\;\begin{array}{cccc}x &=& t + \dfrac{1}{t} & [1] \\ \\[-3mm] y &=& t^2 - \dfrac{1}{t^2} & [2] \end{array}$
$\begin{array}{cccccc}\text{Square [1]:} & x^2 &=& t^2 + 2 + \dfrac{1}{t^2} \\ \\[-3mm] \text{Add [2]:} & y &=& t^2 \;\;-\;\; \dfrac{1}{t^2} \end{array}$

And we have: . $x^2 + y \:=\:2t^2+2 \quad\Rightarrow\quad t^2 \:=\:\frac{x^2+y-2}{2}$

Substitute into $[2]\!:\quad y \;=\;\frac{x^2+y-2}{2} - \frac{2}{x^2+y-2}$

Therefore: . $\boxed{y \;=\;\frac{(x^2+y-2)^2-4}{2(x^2+y-2)}}$

I have no idea what the graph looks like.

I note that: . $x^2+y-2 \:\neq\:0$
That is, the graph does not exist anywhere on the parabola: $y \:=\:2-x^2$

$e)\;\;\begin{array}{cccc}x &=& t + \dfrac{1}{t} & [1] \\ \\[-3mm] y&=& t - \dfrac{1}{t} & [2] \end{array}$
Add [1] and [2]: . $x + y \:=\:2t \quad\Rightarrow\quad t \:=\:\frac{x+y}{2}$

Substitute into [1]: . $x \;=\;\frac{x+y}{2} - \frac{2}{x+y}$

Multiply by $2(x+y)\!:\quad 2x(x+y) \:=\:(x+y)^2-4$

. . which simplifies to: . $\boxed{y^2-x^2 \:=\:4}$

We have a "vertical" hyperbola . $\begin{array}{c}_{\cup} \\ ^{\cap} \end{array}$
. . I'm not sure of the restrictions.

$f)\;\;\begin{array}{cccc}x &=& \dfrac{1 - t^2}{1 + t^2} & [1]\\ \\[-3mm] y &=& \dfrac{2t}{1+t^2} & [2]\end{array}$
I just happen to be familiar with this one . . .

$\begin{array}{ccccc}\text{Square [1]:} & x^2 &=& \dfrac{(1-t^2)^2}{(1+t^2)^2} \\ \\[-3mm]\text{Square [2]:} & y^2 &=& \dfrac{4t^2}{(1+t^2)^2} \end{array}$

Add: . $x^2+y^2 \;=\;\frac{(1-t^2)^2}{(1+t^2)^2} + \frac{4t^2}{(1+t^2)^2} \;=\;\frac{1-2t^2+t^4 + 4t^2}{(1+t^2)^2} \;=\;\frac{1+2t^2+t^4}{(1+t^2)^2}$

. . $x^2+y^2 \;=\;\frac{(1+t^2)^2}{(1+t^2)^2} \quad\Rightarrow\quad \boxed{x^2 + y^2 \:=\:1}$

We have a unit circle centered at the Origin.
. . It has a "hole" at (-1,0). .(Why?)