The transformed equation of 3x^2 - 4xy = r^2 | vedclass

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Question

(C) Let the angle of rotation be $	heta = \operatorname{Tan}^{-1}(2)$,so $	an 	heta = 2$. Then $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta

Vedclass · Accepted Answer

$4y^2 - x^2 = r^2$

Vedclass · Answer

(C) Let the angle of rotation be $	heta = \operatorname{Tan}^{-1}(2)$,so $	an 	heta = 2$. Then $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$.The transformation equations for rotation are $x = x' \cos 	heta - y' \sin 	heta$ and $y = x' \sin 	heta + y' \cos 	heta$.Substituting the values: $x = \frac{x' - 2y'}{\sqrt{5}}$ and $y = \frac{2x' + y'}{\sqrt{5}}$.Substitute these into the given equation $3x^2 - 4xy = r^2$:$3\left(\frac{x' - 2y'}{\sqrt{5}}ight)^2 - 4\left(\frac{x' - 2y'}{\sqrt{5}}ight)\left(\frac{2x' + y'}{\sqrt{5}}ight) = r^2$$\frac{3}{5}(x'^2 - 4x'y' + 4y'^2) - \frac{4}{5}(2x'^2 + x'y' - 4x'y' - 2y'^2) = r^2$$\frac{3}{5}(x'^2 - 4x'y' + 4y'^2) - \frac{4}{5}(2x'^2 - 3x'y' - 2y'^2) = r^2$$\frac{1}{5}(3x'^2 - 12x'y' + 12y'^2 - 8x'^2 + 12x'y' + 8y'^2) = r^2$$\frac{1}{5}(-5x'^2 + 20y'^2) = r^2$$-x'^2 + 4y'^2 = r^2$,which is $4y'^2 - x'^2 = r^2$.

The transformed equation of $3x^2 - 4xy = r^2$ when the coordinate axes are rotated about the origin through an angle of $\operatorname{Tan}^{-1}(2)$ in the positive direction is:

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