The transformed equation of 3x^2 - 6xy + 8y^2 | vedclass

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(B) Let the original coordinates be $(x', y')$ and the new coordinates be $(x, y)$.Given the rotation angle $	heta = \frac{\pi}{4}$,the transformatio

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$5x^2 + 10xy + 17y^2 - 16 = 0$

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(B) Let the original coordinates be $(x', y')$ and the new coordinates be $(x, y)$.Given the rotation angle $	heta = \frac{\pi}{4}$,the transformation equations are:$x' = x \cos \frac{\pi}{4} - y \sin \frac{\pi}{4} = \frac{x - y}{\sqrt{2}}$$y' = x \sin \frac{\pi}{4} + y \cos \frac{\pi}{4} = \frac{x + y}{\sqrt{2}}$Substituting these into the given equation $3(x')^2 - 6x'y' + 8(y')^2 = 8$:$3\left(\frac{x - y}{\sqrt{2}}ight)^2 - 6\left(\frac{x - y}{\sqrt{2}}ight)\left(\frac{x + y}{\sqrt{2}}ight) + 8\left(\frac{x + y}{\sqrt{2}}ight)^2 = 8$$\frac{3}{2}(x^2 + y^2 - 2xy) - \frac{6}{2}(x^2 - y^2) + \frac{8}{2}(x^2 + y^2 + 2xy) = 8$Multiplying by $2$:$3(x^2 + y^2 - 2xy) - 6(x^2 - y^2) + 8(x^2 + y^2 + 2xy) = 16$$(3 - 6 + 8)x^2 + (-6 + 6 + 16)xy + (3 + 6 + 8)y^2 = 16$$5x^2 + 10xy + 17y^2 = 16$$5x^2 + 10xy + 17y^2 - 16 = 0$

The transformed equation of $3x^2 - 6xy + 8y^2 = 8$ when the axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction is:

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