convert parametric to cartesian

[stephc]

u(t)=(t+1/t)i +(t-1/t)j, t>0

show parametric curve above represents the right section of the hyperbola x^2 - y^2 =4 and explain why

please help!!

 

[stapel]

u(t)=(t+1/t)i +(t-1/t)j, t>0

show parametric curve above represents the right section of the hyperbola x^2 - y^2 =4 and explain why

What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

For instance, after reviewing this similar posting (along with the worked examples in your book, in your class notes, and online, such as here), you had:

. . . . .x = t + (1/t)

. . . . .y = t - (1/t)

What did you try next? Please show all of your steps. Thank you!

 

[stephc]

Yes i did that and then I tried squaring both and doing x^2-y^2 and i ended up with x^2-y^2=4/t so I was close but I am not sure what other way to approach the question

 

[Steven G]

Yes i did that and then I tried squaring both and doing x^2-y^2 and i ended up with x^2-y^2=4/t so I was close but I am not sure what other way to approach the question

(t+1/t)^2 is NOT t^2 +2/t + 1/t^2. This should help.

 

[stephc]

thank you!! and im not sure why this curve represents only the right section?

 

[ksdhart]

The equation you were given for u(t) only represents the right-hand side of the hyperbola because the problem specifies that t > 0. The left-hand side of the parabola would be graphed by negative values of t.

 

[soroban]

Hello, stephc!

I'm not sure why this curve represents only the right section.

We have: $\displaystyle \;\begin{Bmatrix}x &=& t + \frac{1}{t} \\ y &=& t - \frac{1}{t} \end{Bmatrix} \;\;{\color{red}{t\,>\,0}}$

Plot the points for $\displaystyle t\,=\,1,2,3,\text { . . .}$
. $\displaystyle \begin{array}{c|cc} t & x & y \\ \hline 1 & 2 & 0 \\ 2 & \frac{5}{2} & \frac{3}{2} \\ 3& \frac{10}{3} & \frac{8}{3} \\ 4 & \frac{17}{4} & \frac{15}{4} \end{array}$

The graph curves upward in Quadrant 1.


Plot the points for $\displaystyle t\,=\,1,\frac{1}{2}, \frac{1}{3},\text{ . . .}$
. . $\displaystyle \begin{array}{c|cc} t & x & y \\ \hline 1 & 2 & 0 \\ \frac{1}{2} & \frac{5}{2} & \text{-}\frac{3}{2} \\ \frac{1}{3} & \frac{10}{3} & \text{-}\frac{8}{3} \\ \frac{1}{4} & \frac{17}{4} & \text{-}\frac{15}{4} \end{array}$

The graph is reflected across the $\displaystyle x$-axis into Quadrant 4.


Therefore, we have the right half of the hyperbola.