The equation of the locus of a point (2 cos t | vedclass

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(D) Let the point be $(x, y) = (2 \cos 	heta-3, 3 \sin 	heta-4)$.From this,we have $\cos 	heta = \frac{x+3}{2}$ and $\sin 	heta = \frac{y+4}{3}$.U

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$9 x^2+4 y^2+54 x+32 y+109=0$

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(D) Let the point be $(x, y) = (2 \cos 	heta-3, 3 \sin 	heta-4)$.From this,we have $\cos 	heta = \frac{x+3}{2}$ and $\sin 	heta = \frac{y+4}{3}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we substitute the expressions:$(\frac{x+3}{2})^2 + (\frac{y+4}{3})^2 = 1$$\frac{x^2+6x+9}{4} + \frac{y^2+8y+16}{9} = 1$Multiplying by $36$ to clear the denominators:$9(x^2+6x+9) + 4(y^2+8y+16) = 36$$9x^2 + 54x + 81 + 4y^2 + 32y + 64 = 36$$9x^2 + 4y^2 + 54x + 32y + 145 - 36 = 0$$9x^2 + 4y^2 + 54x + 32y + 109 = 0$.

The equation of the locus of a point $(2 \cos \theta-3, 3 \sin \theta-4)$ is

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