Respuesta :

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ANSWER

[tex]{x}^{2} + {y}^{2} = 16[/tex]

EXPLANATION

We want to eliminate the parameter from the equation:

[tex]x = 4 \cos(t)...(1)[/tex]

and

[tex]y = 4 \sin(t)...(2)[/tex]

Square both sides of the first equation to get,

[tex] {x}^{2} = (4 \cos(t) )^{2} [/tex]

[tex]{x}^{2} = 16 \cos^{2} (t)...(3)[/tex]

Squaring the second equation gives;

[tex] {y}^{2} = {(4 \sin(t) )}^{2} [/tex]

[tex]{y}^{2} = 16 \sin^{2} (t)...(4)[/tex]

Add equation (3) and (4).

This implies that;

[tex] {x}^{2} + {y}^{2} = 16 \cos^{2} (t) + 16 \sin^{2} (t)[/tex]

Factor the right hand side

[tex] {x}^{2} + {y}^{2} = 16( \cos^{2} (t) + \sin^{2} (t))[/tex]

Recall that;

[tex] \cos^{2} (t) + \sin^{2} (t) = 1[/tex]

[tex] {x}^{2} + {y}^{2} = 16( 1)[/tex]

[tex] {x}^{2} + {y}^{2} = 16[/tex]

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