eliminate parameter of : y= (9cost +4); y=9sint + 1

Question

what is the graph

 

 

hard to simplify 

need help

Answer 1

Hello Pia,

 

Since you have y for both equations I am assuming that this is a set of parametric equations, one of the equations is the parametric representation for x and a function in terms of x, y is what is needed.

If this is true and have chosen the wrong equation for x, just switch the x and y in the final solution.

 

If my assumption is incorrect clarify the problem.  Thank you.

 

Proceeding with this, I will make the equations as follow:

x= 9cost +4

y=9sint + 1

 

For x= 9cost +4 => cos(t) = (x-4)/9

For y=9sint + 1 => sin(t) = (y-1)/9

 

Now cos²(t) + sin²(t) = 1

       (x-4)²/81 + (y-1)²/81 =1

 

∴ (x - 4)² + (y - 1)² = 81

 

This is a circle with center at (4, 1) and radius 9

Answer 2

 

I assume the first statement should be x rather than y.

 

This describes a circle of radius 9 centered at 4, 1.

There are many ways to solve this problem.  For example, using the fact that this is a circle, we can write the solution as:

 

y = sqrt(81 - (x - 4)²) + 1

y = -sqrt(81 - (x - 4)²) + 1

 

Another way to solve this would be to use the fact that sin = sqrt(1 - cos²) and first reparametrize as

 

x = 9 t + 4

y = 9 sqrt(1-t²) + 1, y = -9 sqrt(1-t²) + 1

 

Then, we may solve for t in the first expression as

t = (x-4)/9

Substituting,

y = 9 sqrt(1-((x-4)/9)²) + 1, y = -9 sqrt(1-((x-4)/9)²) + 1

Moving the 9 outside the square root to 81 inside, we get

 

y = sqrt(81 - (x - 4)²) + 1
y = -sqrt(81 - (x - 4)²) + 1

This is the same as above.