Converting parametric equations into a Cartesian equation

[Weng123]

I understand that to convert parametric equations into a Cartesian equation, we can solve for the third variable (in this case; t) in terms of x and then plug that into the other equation thus giving you an equation with only x and y as the variable.
However for this particular question, because trigonometric functions are involved there is another way to convert the parametric equations into a Cartesian equation. I understand how to solve for the answer but I just don’t understand why the first method doesn’t also yield the correct answer. I have taken a picture of the question with my work below.
Thank you very much!

 

[Deleted member 4993]

I understand that to convert parametric equations into a Cartesian equation, we can solve for the third variable (in this case; t) in terms of x and then plug that into the other equation thus giving you an equation with only x and y as the variable.
However for this particular question, because trigonometric functions are involved there is another way to convert the parametric equations into a Cartesian equation. I understand how to solve for the answer but I just don’t understand why the first method doesn’t also yield the correct answer. I have taken a picture of the question with my work below.View attachment 24739
Thank you very much!

How do you know the "alternative" method is providing wrong answer!!

 

[Weng123]

How do you know the "alternative" method is providing wrong answer!!

Because in the answer key, the correct answer is the equation given by the second method.

 

[pka]

Because in the answer key, the correct answer is the equation given by the second method.

Well the standard for an ellipse is $\large \dfrac{(x-h)^2}{a^2}+\dfrac{(y+k)^2}{b^2}=1$
where the centre is $(h,k)$. Now why do you think there any other way?

 

[Steven G]

Because in the answer key, the correct answer is the equation given by the second method.

That is not a good reason at all to say that your answer is wrong. There are always different ways of writing the same equation. It is standard to solve the problem using the Correct Way you posted but one does not need to be standard.

Look at the graph of both and see for yourself if they are the same or not.


3x+4y = 12 and y = -3x/4 + 3 look different but are the same!

 

[Weng123]

That is not a good reason at all to say that your answer is wrong. There are always different ways of writing the same equation. It is standard to solve the problem using the Correct Way you posted but one does not need to be standard.

Look at the graph of both and see for yourself if they are the same or not.


3x+4y = 12 and y = -3x/4 + 3 look different but are the same!

Yes, that’s why I graphed both answers out on the right and they are different.

 

[Steven G]

Yes, that’s why I graphed both answers out on the right and they are different.

Yes, you did graph both equations.

Here is the problem. The graph of the correct way is NOT a graph of a function (it fails the vertical line test).
Your solution is a function. You should have put y = +/- 5cos(arcsin(x/4) to get the complete answer.

 

[Weng123]

Yes, you did graph both equations.

Here is the problem. The graph of the correct way is NOT a graph of a function (it fails the vertical line test).
Your solution is a function. You should have put y = +/- 5cos(arcsin(x/4) to get the complete answer.

I realize that the correct way is not a graph of a function however the question never said that the equation we had to find is a function. It just states for us to convert the given parametric functions in a Cartesian equation. Moreover I don’t think it is possible for us to add +/- in front of an equation just like that.

 

[Steven G]

I realize that the correct way is not a graph of a function however the question never said that the equation we had to find is a function. It just states for us to convert the given parametric functions in a Cartesian equation. Moreover I don’t think it is possible for us to add +/- in front of an equation just like that.

If you do not restrict the domain of the arcsin, then you do get plus/minus.

You arrived at t = arcsin(x/4). Now for any x, you get one solution for t. However, if you solve for t in sin(t) = x/4 (lets assume that t is in [0, 2pi]), then you do get two answers for t--one in quadrant 1 and a second result in angle 2. That 2nd solution is y = 5cos[pi - arcsin(x/4)] = -5cos(arcsin(x/4)).
I used the identity that cos(pi - x) = -cos(x)!!

Is this all clear now.

 

[HallsofIvy]

Since $\displaystyle sin^2(\theta)+ cos^2(\theta)= 1$, $\displaystyle cos(\theta)= \pm\sqrt{1- sin^2(\theta)}$. Taking $\displaystyle \theta= arcsin(x/4)$ that becomes $\displaystyle cos(arcsin(x/4))= \pm\sqrt{1- sin^2(arcsin(x/4)}$.

Of course, $\displaystyle sin(arcsin(x/4))= x/4$ so that $\displaystyle cos(arcsin(x/4))= \pm\sqrt{1- \frac{x^2}{16}}= \pm\sqrt{\frac{16- x^2}{16}}= \frac{\pm\sqrt{16- x^2}}{4}$.