Parametric Q (1 Viewer)

[CriminalCrab]

eliminate parameters and hence find cartesian equation:
x=t+(1/t) and y= t^2+(1/t^2)

answer: x^2 -2
what shape is it?

 

[tohriffic]

Sorry I haven't a clue but you can check out http://www.geogebra.org/webstart/geogebra.html, it's a great graph sketching tool!

 

[SpiralFlex]

This graph is called a hyperbola. Commonly used in 4U Conics. As for your parametric equations, I am sure you typed the question/answer incorrectly. Picture a "smiley face above a sad face." You will learn how to find the foci, eccentricity, asymptotes and directrices later in your study of 4U mathematics.

 

[barbernator]

This graph is called a hyperbola. Commonly used in 4U Conics. As for your parametric equations, I am sure you typed the question/answer incorrectly. Picture a "smiley face above a sad face." You will learn how to find the foci, eccentricity, asymptotes and directrices later in your study of 4U mathematics.

the answer to your question is actually y=x(2-x)

note: this isn't the same as what 3U students call a hyperbola

 

[bleakarcher]

eliminate parameters and hence find cartesian equation:
x=t+1/t and y= t^2+1/t^2

answer: 4(y+4)^2 - 9(x-1)^2 = 36
what shape is it?

y=t^2+(1/t)^2=[t+(1/t)]^2-2=x^2-2
I dont get it?

 

[SpiralFlex]

the answer to your question is actually y=x(2-x)

note: this isn't the same as what 3U students call a hyperbola

Yes, that's what I said. So it must be the answer or question that is wrong.

 

[barbernator]

y=t^2+(1/t)^2=[t+(1/t)]^2-2=x^2-2
I dont get it?

i think the question is grouped differently, i.e. x=(t+1)/t and y=(t^2+1)/(t^2)

 

[bleakarcher]

OP, use parentheses lol

 

[xXnukerrrXx]

dunno if this would help?

(1) x=t+(1/t)
(2) y=t^2+(1/t^2)

similarly to the property a^2+b^2=(a+b)^2-2ab
(1) x^2=(t+(1/t))^2=t^2+(1/t^2)+2
therefore to get t^2+(1/t^2) we minus the 2
x^2-2=t^2+(1/t^2)
substitute t^2+(1/t^2) in equation (2)
y=x^2-2

 

[CriminalCrab]

yea that s how its meant to be grouped sorry, and i read answer wrong too, argh
will fix
anyway, its in the 3u cambridge textbook

 

[Aesytic]

i thought this was still 3U, and it was only 4U when it involved other graphs like ellipses and hyperbolas :/

 

[SpiralFlex]

The solution provided by CriminalCrab is 4U. However we believe he typed it incorrectly.

 

[cutemouse]

Isn't that a parabola? ...

 

[SpiralFlex]

Isn't that a parabola? ...

Correct, so I would assume that the answer the OP provided us is wrong.

 

[viraj30]

x= t+(1/t)..

therefore x^2= t^2+2+(1/t^2)

so, x^2-2=t^2+(1/t^2)

...y=x^2-2

EASY!

 

[cutemouse]

Also, it isn't all of the parabola. It's the part of parabola for which x <= -2 or x>=2, and y >= 2...