Converting Trig Parametric equations for ellipse to cartesian equation

[H.Bisho18]

HI

I have this parametric equation problem.


and im not sure what to do since its cos(t+pie/4) and sin(t). Normally i would rearrange and use the sin^2 + cos^2 =1 identity but that wont work here.
I also tried using the double angle cos rules but i got lost from there.

thanks for any help

 

[MarkFL]

Hello, and welcome to FMH!

Can you verify that:

[MATH]x+y=2\sqrt{3}\cos(t)[/MATH] ?

 

[H.Bisho18]

Hello, and welcome to FMH!

Can you verify that:

[MATH]x+y=2\sqrt{3}\cos(t)[/MATH] ?

Thanks for the quick reply, and no im not sure how to do that.
Is it just sub to 2 trigs in and simplify it down?

 

[Deleted member 4993]

HI

I have this parametric equation problem.
View attachment 11662

and im not sure what to do since its cos(t+pie/4) and sin(t). Normally i would rearrange and use the sin^2 + cos^2 =1 identity but that wont work here.
I also tried using the double angle cos rules but i got lost from there.

thanks for any help

Do you know a general equation (in cartesian co-ordinates) for an oblique ellipse?

 

[H.Bisho18]

Do you know a general equation (in cartesian co-ordinates) for an oblique ellipse?

I dont. I just know the dy/dx=dy/dt * dt/x and the using trig identities to get the cartesian

 

[MarkFL]

Thanks for the quick reply, and no im not sure how to do that.
Is it just sub to 2 trigs in and simplify it down?

Using the angle-sum identity for cosine, can you rewrite:

[MATH]4\cos\left(t+\frac{\pi}{6}\right)[/MATH] ?

 

[H.Bisho18]

Using the angle-sum identity for cosine, can you rewrite:

[MATH]4\cos\left(t+\frac{\pi}{6}\right)[/MATH] ?

Yep 4(cos(t)cos(π/6) -sin(t)sin(π/6))

i got that then tried to rearrange the y=2sin(t) for t and sub it in. but i didnt get anywhere

 

[MarkFL]

Yep 4(cos(t)cos(π/6) -sin(t)sin(π/6))

i got that then tried to rearrange the y=2sin(t) for t and sub it in. but i didnt get anywhere

The angle [MATH]\frac{\pi}{6}[/MATH] is one of our "special angles" for which we know the value of the trig. functions:

[MATH]\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}[/MATH]
[MATH]\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}[/MATH]
Using these, we then find:

[MATH]4\cos\left(t+\frac{\pi}{6}\right)=4\left(\frac{\sqrt{3}}{2}\cos(t)-\frac{1}{2}\sin(t)\right)=2\sqrt{3}\cos(t)-2\sin(t)[/MATH]
And so:

[MATH]x+y=\left(2\sqrt{3}\cos(t)-2\sin(t)\right)+\left(2\sin(t)\right)=2\sqrt{3}\cos(t)[/MATH]
So now, consider the sum:

[MATH](x+y)^2+ay^2[/MATH]
Can you see that this may be written as:

[MATH](x+y)^2+ay^2=12\cos^2(t)+4a\sin^2(t)[/MATH] ?

 

[H.Bisho18]

The angle [MATH]\frac{\pi}{6}[/MATH] is one of our "special angles" for which we know the value of the trig. functions:

[MATH]\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}[/MATH]
[MATH]\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}[/MATH]
Using these, we then find:

[MATH]4\cos\left(t+\frac{\pi}{6}\right)=4\left(\frac{\sqrt{3}}{2}\cos(t)-\frac{1}{2}\sin(t)\right)=2\sqrt{3}\cos(t)-2\sin(t)[/MATH]
And so:

[MATH]x+y=\left(2\sqrt{3}\cos(t)-2\sin(t)\right)+\left(2\sin(t)\right)=2\sqrt{3}\cos(t)[/MATH]
So now, consider the sum:

[MATH](x+y)^2+ay^2[/MATH]
Can you see that this may be written as:

[MATH](x+y)^2+ay^2=12\cos^2(t)+4a\sin^2(t)[/MATH] ?

ah yes that makes alot of sense now. thank you so much