Precal help.
A pair of parametric equations is given.
x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π

Question

Precal help.
A pair of parametric equations is given.
x = 2 cos t,    y = 3 sin t,    0 ≤ t ≤ 2π

anonymous · Answer

(a) Sketch the curve represented by the parametric equations.
It is an elipse
(b) Find a rectangular-coordinate equation for the curve by eliminating the parameter. 
The answer is $$x^2+\frac{ 4 }{9 }y^2=4$$

How do you get this?

ganeshie8 · Answer

x = 2 cos t

=>

x/2 = cost

ganeshie8 · Answer

y = 3 sin t

=>

y/3  = sint

ganeshie8 · Answer

next pull the well known trig identity :sin^2t + cos^2t = 1

ganeshie8 · Answer

(x/2)^2 + (y/3)^2 = 1

ganeshie8 · Answer

simplify

anonymous · Answer

Thats not exactly how I got mine

x=2cost, y=3sint

square both sides
x^2=4cos^2t
y^2=9sin^2t

y^2=9(1-cos^2t)
y^2=9-9cos^2t
((y^2-9)/-9)=cos^2t

I get x^2=4((9+y^2)/9)

Now what I don't get is how they get this to

$$x^2= 4-\frac{ 4 }{ 9 }y^2$$

which turns into

$$x^2+\frac{ 4 }{ 9 }y^2=4$$

ganeshie8 · Answer

there was a mistake in ur calculation

ganeshie8 · Answer

Thats not exactly how I got mine

x=2cost, y=3sint

square both sides
x^2=4cos^2t
y^2=9sin^2t

y^2=9(1-cos^2t)
y^2=9-9cos^2t
((y^2-9)/-9)=cos^2t

I get x^2=4((9+y^2)/9) **********************************

Now what I don't get is how they get this to

ganeshie8 · Answer

that line,
how come (y^2-9)/-9   turned into  (9+y^2)/9 ??

anonymous · Answer

Negative divided by negative = positive right?
This is a method used by my teacher, so I never completely understood how he got that.

ganeshie8 · Answer

lets see :)

you have :

 $\large  \frac{y^2-9}{-9}$

ganeshie8 · Answer

multiply top and bottom wid "-1"

ganeshie8 · Answer

$\large  \frac{(y^2-9)\times (-1)}{-9\times  (-1)}  $

ganeshie8 · Answer

$\large  \frac{-y^2+9}{9}  $

ganeshie8 · Answer

$\large  \frac{9-y^2}{9}  $

anonymous · Answer

Oh Sorry that was what he put, the 9-y^2/9

anonymous · Answer

So now how does that turn into 
$$4-\frac{ 4 }{ 9 }y^2$$

ganeshie8 · Answer

so, you're fine till  below line :

$\large     x^2=4\left(\frac{9-y^2}{9}\right)$

?

anonymous · Answer

Yeah, I don't get how it turns to what I previously showed.

ganeshie8 · Answer

first divide the 9 inside parenthesis

ganeshie8 · Answer

$\large     x^2=4\left(\frac{9-y^2}{9}\right)$


$\large     x^2=4\left(\frac{9}{9} -  \frac{y^2}{9}\right)$

ganeshie8 · Answer

still fine ?

anonymous · Answer

So that would get you $$x^2=4(1-\frac{ y^2 }{ 9})$$
=4-4y^2/9?

anonymous · Answer

AH alright, that makes total sense. Thank you so much man.

ganeshie8 · Answer

np.. u wlc :)