Question

Given x = 2sin (nt + pi/3) and y= 4sin (nt + pi/6),
Find the Cartesian equation of the locus of the point (x,y) as t varies.

Answer 1

I've already expressed x and y in terms of sin nt and cos nt:

\[X = \sin nt + \sqrt 3 \cos nt\]

\[Y= 2\sqrt3 \sin nt + 2 \cos nt\]

I'm not exactly sure as to how the question wants me to proceed.

Thanks in advance!

Answer 2

hint:
we can apply this identity:
\[\begin{gathered}
x = 2\sin \left( {nt + \frac{\pi }{3}} \right) = 2\sin \left( {nt + \frac{\pi }{6} + \frac{\pi }{6}} \right) = 2\sin \left( {\left( {nt + \frac{\pi }{6}} \right) + \frac{\pi }{6}} \right) = \hfill \\
\hfill \\
= 2\sin \left( {nt + \frac{\pi }{6}} \right)\cos \left( {\frac{\pi }{6}} \right) + 2\cos \left( {nt + \frac{\pi }{6}} \right)\sin \left( {\frac{\pi }{6}} \right) = \hfill \\
\hfill \\
= 2 \cdot \frac{y}{4} \cdot \cos \left( {\frac{\pi }{6}} \right) + 2 \cdot \sqrt {1 - {{\left( {\frac{y}{4}} \right)}^2}} \cdot \sin \left( {\frac{\pi }{6}} \right) \hfill \\
\end{gathered} \]

Answer 3

Thanks for the hint! I'll look at some more videos and see if I can solve it!

Answer 4

@Michele_Laino

I'm sorry to bother you, but I'm not how to deal with this.

My book says: \[4x^2 + y^2 - 2\sqrt3xy = 4\]

I worked backwards from the answer, and it seems to simplify to: \[12\sin^2 + 12 \cos^2 = 4\]

I don't know where the 4 comes from at all!

Answer 5

here is the next step:
\[\begin{gathered}
x = 2\sin \left( {nt + \frac{\pi }{3}} \right) = 2\sin \left( {nt + \frac{\pi }{6} + \frac{\pi }{6}} \right) = 2\sin \left( {\left( {nt + \frac{\pi }{6}} \right) + \frac{\pi }{6}} \right) = \hfill \\
\hfill \\
= 2\sin \left( {nt + \frac{\pi }{6}} \right)\cos \left( {\frac{\pi }{6}} \right) + 2\cos \left( {nt + \frac{\pi }{6}} \right)\sin \left( {\frac{\pi }{6}} \right) = \hfill \\
\hfill \\
= 2 \cdot \frac{y}{4} \cdot \cos \left( {\frac{\pi }{6}} \right) + 2 \cdot \sqrt {1 - {{\left( {\frac{y}{4}} \right)}^2}} \cdot \sin \left( {\frac{\pi }{6}} \right) = \hfill \\
\hfill \\
= \frac{y}{2} \cdot \frac{{\sqrt 3 }}{2} + \frac{{\sqrt {4 - {y^2}} }}{2} \hfill \\
\end{gathered} \]
so we can write:
\[x = \frac{y}{2} \cdot \frac{{\sqrt 3 }}{2} + \frac{{\sqrt {4 - {y^2}} }}{2}\]

Answer 6

So:

\[4x^2 = 2y^2 +2\sqrt (3y - y^3) +4 \]

Answer 7

sorry, I have made a typo, here is the right final step:
\[x = \frac{y}{2} \cdot \frac{{\sqrt 3 }}{2} + \frac{{\sqrt {16 - {y^2}} }}{4}\]

Answer 8

Ah! let me rework it then.

Answer 9

and we can rewrite such equation, like this:
\[x - \frac{y}{2} \cdot \frac{{\sqrt 3 }}{2} = \frac{{\sqrt {16 - {y^2}} }}{4}\]

Answer 10

please square both sides

Answer 11

I'm squaring it now, just bear with me please!

Answer 12

hint:
\[{\left( {x - \frac{y}{2} \cdot \frac{{\sqrt 3 }}{2}} \right)^2} = {\left( {\frac{{\sqrt {16 - {y^2}} }}{4}} \right)^2}\]

Answer 13

\[x^2 - \frac{ \sqrt3 xy }{ 2 } + \frac{ 3y^2 }{ 16 } = \frac{16 - y^2? }{ 16 } \]

Answer 14

correct!

Answer 15

therefore:

\[16x^2 - 8\sqrt3 xy +3y^2 = 16 - y^2 = 16x^2 - 8\sqrt3 xy + 4y^2 = 16\]

Answer 16

\[4x^2 - 2\sqrt3 xy + y ^2 = 4\]

Answer 17

Ah! thank you so much for your help. I see how my book came to the answer!

Answer 18

:)

Answer 19

Alright, I'll be closing the question now. Thanks once again for your help and patience!

Answer 20

:)