cartesian equation of the surface with parametric equation

[wiloben25]

Hi all, the question is follows.

Find the cartesian equation of the surface with parametric specification and identify the surface.
x=8cosh(theta)cos(phi)
y=13sinh(theta)cos(phi)
z=3sin(phi)

i have squared everything which gives
x²/64=cosh²(theta)cos²(phi)
y²/169=sinh²(theta)cos²(phi)
z²/9=sin²(phi)

Am i correct in using the fact that this would be x²/64+y²/169+z²/9 and letting it equal 1, which produces an elipsoid, or have i completely stuffed this one up

Thanks

 

[Deleted member 4993]

Hi all, the question is follows.

Find the cartesian equation of the surface with parametric specification and identify the surface.
x=8cosh(theta)cos(phi)
y=13sinh(theta)cos(phi)
z=3sin(phi)

i have squared everything which gives
x²/64=cosh²(theta)cos²(phi)
y²/169=sinh²(theta)cos²(phi)
z²/9=sin²(phi)
Am i correct in using the fact that this would be x²/64+y²/169+z²/9 and letting it equal 1, which produces an elipsoid, or have i completely stuffed this one up

Thanks

cosh²(Θ) + sinh²(Θ) = ? and

cosh²(Θ) - sinh²(Θ) = ?

 

[wiloben25]

cosh²(Θ) + sinh²(Θ) = ? and

cosh²(Θ) - sinh²(Θ) = ?

cosh²(Θ) + sinh²(Θ) = cosh2(Θ) and cosh²(Θ) - sinh²(Θ) = 1?

but where does the minus come from?

 

[Deleted member 4993]

cosh²(Θ) + sinh²(Θ) = cosh2(Θ) and cosh²(Θ) - sinh²(Θ) = 1?

but where does the minus come from?

From the definitions of cosh(Θ) and sinh(Θ)

[Just like you can show

cos²(Θ) + sin²(Θ) = 1

from their respective definitions]

 

[Deleted member 4993]

Hi all, the question is follows.

Find the cartesian equation of the surface with parametric specification and identify the surface.
x=8cosh(theta)cos(phi)
y=13sinh(theta)cos(phi)
z=3sin(phi)

i have squared everything which gives
x²/64=cosh²(theta)cos²(phi)
y²/169=sinh²(theta)cos²(phi)
z²/9=sin²(phi)

Am i correct in using the fact that this would be x²/64+y²/169+z²/9 and letting it equal 1, which produces an elipsoid, or have i completely stuffed this one up

Thanks

cosh²(Θ) - sinh²(Θ) = 1

Rectify your equation in view of the above stated relationship.

 

[wiloben25]

ok so then...

z²/9+sin²(phi)=x²/64+cosh²(theta)cos²(phi)+y²/169+sinh²(theta)cos²(phi)
z²/9=x²/64+y²/169+2cosh(Θ)+2cos²(Θ)-sin²(phi)?

I dont think this correct and its where im breaking down/confused

 

[wiloben25]

z²/9=x²/64+y²/169+2cosh(Θ)+2cos²(Θ)-sin²(phi)?

or does this somehow get to equaling 2+cos²(phi)
so z²/9=x²/64+y²/169+2+cos²(phi)

 

[Deleted member 4993]

Hi all, the question is follows.

Find the cartesian equation of the surface with parametric specification and identify the surface.
x=8cosh(theta)cos(phi)
y=13sinh(theta)cos(phi)
z=3sin(phi)

i have squared everything which gives
x²/64=cosh²(theta)cos²(phi)
y²/169=sinh²(theta)cos²(phi)
z²/9=sin²(phi)

Am i correct in using the fact that this would be x²/64+y²/169+z²/9 and letting it equal 1, which produces an elipsoid, or have i completely stuffed this one up

Thanks

(x/8)^2 - (y/13)^2 + (z/3)^2 = ?