When the coordinate axes are rotated by an an | vedclass

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Question

(C) When the coordinate axes are rotated by an angle $	heta$ about the origin,the transformation equations are given by:$x = x_1 \cos 	heta - y_1 \s

Vedclass · Accepted Answer

$x^2+y^2=9$

Vedclass · Answer

(C) When the coordinate axes are rotated by an angle $	heta$ about the origin,the transformation equations are given by:$x = x_1 \cos 	heta - y_1 \sin 	heta$$y = x_1 \sin 	heta + y_1 \cos 	heta$Given $	heta = 	an ^{-1} \left(\frac{3}{4}ight)$,we have $	an 	heta = \frac{3}{4}$.From the right-angled triangle with opposite side $3$ and adjacent side $4$,the hypotenuse is $\sqrt{3^2+4^2} = 5$.Thus,$\cos 	heta = \frac{4}{5}$ and $\sin 	heta = \frac{3}{5}$.Substituting these into the transformation equations:$x = \frac{4}{5} x_1 - \frac{3}{5} y_1$$y = \frac{3}{5} x_1 + \frac{4}{5} y_1$Substitute these into the original equation $x^2 + y^2 = 9$:$\left(\frac{4}{5} x_1 - \frac{3}{5} y_1ight)^2 + \left(\frac{3}{5} x_1 + \frac{4}{5} y_1ight)^2 = 9$$\frac{1}{25} \left( (4x_1 - 3y_1)^2 + (3x_1 + 4y_1)^2 ight) = 9$$\frac{1}{25} \left( 16x_1^2 + 9y_1^2 - 24x_1y_1 + 9x_1^2 + 16y_1^2 + 24x_1y_1 ight) = 9$$\frac{1}{25} \left( 25x_1^2 + 25y_1^2 ight) = 9$$x_1^2 + y_1^2 = 9$Thus,the transformed equation is $x^2 + y^2 = 9$.

When the coordinate axes are rotated by an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ about the origin,then the equation $x^2+y^2=9$ is transformed to the equation.

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