Thanks for the quick reply, and no im not sure how to do that.Hello, and welcome to FMH!
Can you verify that:
[MATH]x+y=2\sqrt{3}\cos(t)[/MATH] ?
Do you know a general equation (in cartesian co-ordinates) for an oblique ellipse?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?
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?
Yep 4(cos(t)cos(π/6) -sin(t)sin(π/6))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
ah yes that makes alot of sense now. thank you so muchThe 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] ?