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

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(C) Given $	heta = 	an^{-1}(2)$,so $	an 	heta = 2$. Since $	an 	heta = \frac{2}{1}$,we have $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta

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$4Y^2 - X^2 = r^2$

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(C) Given $	heta = 	an^{-1}(2)$,so $	an 	heta = 2$. Since $	an 	heta = \frac{2}{1}$,we have $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$. The transformation equations for rotation of axes 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 original equation $3x^2 - 4xy = r^2$: $3\left(\frac{X - 2Y}{\sqrt{5}}ight)^2 - 4\left(\frac{X - 2Y}{\sqrt{5}}ight)\left(\frac{2X + Y}{\sqrt{5}}ight) = r^2$. $\frac{3}{5}(X^2 - 4XY + 4Y^2) - \frac{4}{5}(2X^2 + XY - 2Y^2) = r^2$. $\frac{1}{5}(3X^2 - 12XY + 12Y^2 - 8X^2 - 4XY + 8Y^2) = r^2$. $\frac{1}{5}(-5X^2 - 16XY + 20Y^2) = r^2$. Wait,re-evaluating the original equation $3x^2 - 4xy = r^2$ with rotation $	heta = 	an^{-1}(2)$: Using the general form $ax^2 + 2hxy + by^2 = r^2$,where $a=3, h=-2, b=0$. The new coefficients $a', h', b'$ are given by $a' = a \cos^2 	heta + 2h \sin 	heta \cos 	heta + b \sin^2 	heta$ and $b' = a \sin^2 	heta - 2h \sin 	heta \cos 	heta + b \cos^2 	heta$. $a' = 3(\frac{1}{5}) + 2(-2)(\frac{2}{5}) + 0 = \frac{3-8}{5} = -1$. $b' = 3(\frac{4}{5}) - 2(-2)(\frac{2}{5}) + 0 = \frac{12+8}{5} = 4$. $h' = (b-a) \sin 	heta \cos 	heta + h(\cos^2 	heta - \sin^2 	heta) = (0-3)(\frac{2}{5}) + (-2)(\frac{1-4}{5}) = -\frac{6}{5} + \frac{6}{5} = 0$. Thus,the equation becomes $-X^2 + 4Y^2 = r^2$,or $4Y^2 - X^2 = r^2$.

The transformed equation of $3x^2 - 4xy = r^2$ when the coordinate axes are rotated through an angle $\tan^{-1}(2)$ is:

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