Question

HELP NEEDED PLEASE!!!

Eliminate the parameter. x = 4 cos t, y = 4 sin t

Answer 1

@mathstudent55 @SolomonZelman @zepdrix

Answer 2

I'm kind of new to these also. Perhaps you could solve each for t and set them equal to one another?

Answer 3

x=4cost => cost=x/4 and y=4sint => sint=y/4 now x^2/16+y^2/16 = cos^2t+sin^2t = 1

Answer 4

So \[\cos^{-1}(\frac{x}{4}) = \sin^{-1}(\frac{y}{4})\]

Answer 5

sorry: x^2/16+y^2/16+xy/8=1

Answer 6

um..so \[\frac{ x ^{2} }{ 16}+\frac{ y^{2} }{ 16}+\frac{ xy }{ 8}=1\]

Answer 7

I think I'm starting to fade. :P

Answer 8

im just so confused

Answer 9

Give me another minute, I think I'm almost there.

Answer 10

\[\large{\cos^2 t + \sin^2 t = 1}\]

Answer 11

\[\large{x = 4\cos t}\]

\[\large{\implies \cos t = \cfrac{x}{4}}\]

Similarly,

\[\large{\sin t = \cfrac{y}{4}}\]

Answer 12

Ah duh! Thanks vishweshrimali5! That's it. :)

Answer 13

Substitute these values of cos t and sin t in :

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
\[\large{\cos^2 t + \sin^2 t = 1}\]
\(\color{blue}{\text{End of Quote}}\)

And you will get your answer :)

Answer 14

thank you :)

Answer 15

Your welcome :)