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    I'm stuck at part c.. I just looked at the solution and it used sin²x + cos²x = 1 to solve it but I don't understand why my way is giving the wrong answer!! I found the length of the line using point C which gives the radius and the center is (0,0) --> (x)² + (y)² = 57/4

    The correct answer is (x/4)² + (y/3)² =1 .. Can someone pls tell me where I went wrong, thanks
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    Hey! I did one like this the other day

    What you have to do is make the x equation cos(t)= and the y equation sin(t)=.
    So cos(t)= (x/4) and sin(t)= (y/3)
    So then you have sin and cos!

    Next you will need to square them, because we will need to make an equation involving x and y, and we know this is an equation of a circle. I usually just assume that if you have sin and cos, you will be using that c2 trig identity.

    So you will have cos^2(t) = (x/4)^2 and sin^2(t) = (y/3)^2
    Then - in the trig equation - you can replace the cos^2 and the sin^2 with the x and y fractions.

    I think the logic of your way does make sense - did you calculate the length of the line from point c to the centre? You might want to double check your calculation and your answer for part B

    I hope this helped!
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    (Original post by pondsteps)
    I'm stuck at part c.. I just looked at the solution and it used sin²x + cos²x = 1 to solve it but I don't understand why my way is giving the wrong answer!! I found the length of the line using point C which gives the radius and the center is (0,0) --> (x)² + (y)² = 57/4

    The correct answer is (x/4)² + (y/3)² =1 .. Can someone pls tell me where I went wrong, thanks
    You went wrong because this is not a circle, it's an ellipse.

    Circles are in the parametric form x=a\cos(t) + g\quad y=a\sin(t) + h, \quad t\in [0,2\pi), \quad \forall a, g ,h \in \mathbb{R} where the centre is (g,h) and the radius is a. An ellipse is defined in such a way where the coefficients of \cos(t) and \sin(t) are NOT the same.

    So since it is an ellipse and not a circle, the way you have defined the equation requires every point to be equidistant from the centre point which is the definition of a circle, and in your case your distance of every point was the length from O to C - and this is not the case for any ellipse as all the points are not equidistant from the centre..
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    a lot of these rely on the identity

    sin2Θ+ cos2Θ ≡ 1
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    (Original post by RDKGames)
    You went wrong because this is not a circle, it's an ellipse.

    Circles are in the parametric form x=a\cos(t) + g\quad y=a\sin(t) + h, \quad t\in [0,2\pi), \quad \forall a, g ,h \in \mathbb{R} where the centre is (g,h) and the radius is a. An ellipse is defined in such a way where the coefficients of \cos(t) and \sin(t) are NOT the same.

    So since it is an ellipse and not a circle, the way you have defined the equation requires every point to be equidistant from the centre point which is the definition of a circle, and in your case your distance of every point was the length from O to C - and this is not the case for any ellipse as all the points are not equidistant from the centre..
    Ohhhh omg thank u so much
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    (Original post by ~Martha~)
    Hey! I did one like this the other day

    What you have to do is make the x equation cos(t)= and the y equation sin(t)=.
    So cos(t)= (x/4) and sin(t)= (y/3)
    So then you have sin and cos!

    Next you will need to square them, because we will need to make an equation involving x and y, and we know this is an equation of a circle. I usually just assume that if you have sin and cos, you will be using that c2 trig identity.

    So you will have cos^2(t) = (x/4)^2 and sin^2(t) = (y/3)^2
    Then - in the trig equation - you can replace the cos^2 and the sin^2 with the x and y fractions.

    I think the logic of your way does make sense - did you calculate the length of the line from point c to the centre? You might want to double check your calculation and your answer for part B

    I hope this helped!
    Thanks a bunch!!!
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    (Original post by the bear)
    a lot of these rely on the identity

    sin2Θ+ cos2Θ ≡ 1
    Thank u!
 
 
 
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