Need help with a parametric equation question thanks (1 Viewer)

[Bigboigood]

This question has me pretty stumped. It's question 14 d) from the Sydney Girls High School 2020 Maths Extension 1 Paper:
Find the cartesian equation of the curve with the parametric equations:
x=4cos(theta)-sin(theta) y=5cos(theta)+sin(theta)

Sorry for the terrible formatting and thanks in advance.

 

[Greninja340]

could you just solve both of them for sin(theta) and then equate them and solve for y

 

[Bigboigood]

There would still be a cos(theta) in the equation wouldn't there?

 

[fan96]

Observe that





so

 

[cossine]

x = 4*cos(theta) - sin(theta)

Use the theorem:
cos(theta + phi) = cos(theta)*cos(phi) - sin(theta)*sin(phi)

=> r*cos(theta + phi) = r*cos(theta)*cos(phi) - r*sin(theta)*sin(phi)

Equate

x = r*cos(theta)*cos(phi) - r*sin(theta)*sin(phi)

=> r*cos(phi) = 4, r*sin(phi) =1

Therefore (r^2)*cos^2(phi) + (r^2)*sin^2(phi) = 17

Using theorem: sin^2(theta) + cos^2(theta) = 1

=> r = sqrt(17)

tan(phi) = (r*sin(phi))/ (r*cos(phi) ) = 1/4


Using theorem (f^-1( f(x) ) = x ):

=> phi = tan^-1(1/4)


Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))

=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta

Substitute the expression for theta into y

 

[fan96]

Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))

=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta

Substitute the expression for theta into y

If you plot this using software the graph you get is not the same as the original parametric equation.
(You only get the upper half of the graph.)

Using theorem (f^-1( f(x) ) = x ):
...
Therefore x = sqrt(17)*cos(theta + tan^-1(1/4))
=> cos^-1( x/sqrt(17) ) - tan^-1(1/4) = theta

This theorem requires to be invertible wherever is defined.
The input varies over an interval of length , where does not have an inverse.

Perhaps an easier way to state this is that if then the range of is an interval of length , yet to get the entire graph of the curve we want we must take over a range of length at least .

This problem is fixed by adding a in front, so

 

[cossine]

If you plot this using software the graph you get is not the same as the original parametric equation.
(You only get the upper half of the graph.)


This theorem requires to be invertible wherever is defined.
The input varies over an interval of length , where does not have an inverse.

Perhaps an easier way to state this is that if then the range of is an interval of length , yet to get the entire graph of the curve we want we must take over a range of length at least .

This problem is fixed by adding a in front, so

Thank you