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(B) When the coordinate axes are rotated by an angle $	heta = \frac{\pi}{4}$ in the positive direction,the transformation equations are: $x = X \cos

Vedclass · Accepted Answer

$9$

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(B) When the coordinate axes are rotated by an angle $	heta = \frac{\pi}{4}$ in the positive direction,the transformation equations are: $x = X \cos 	heta - Y \sin 	heta = \frac{X-Y}{\sqrt{2}}$ $y = X \sin 	heta + Y \cos 	heta = \frac{X+Y}{\sqrt{2}}$ Substituting these into the equation $25x^2 + 9y^2 = 225$: $25 \left( \frac{X-Y}{\sqrt{2}} ight)^2 + 9 \left( \frac{X+Y}{\sqrt{2}} ight)^2 = 225$ $\frac{25}{2} (X^2 + Y^2 - 2XY) + \frac{9}{2} (X^2 + Y^2 + 2XY) = 225$ $\frac{34X^2 + 34Y^2 - 32XY}{2} = 225$ $17X^2 - 16XY + 17Y^2 = 225$ Comparing this with $\alpha x^2 + \beta xy + \gamma y^2 = \delta$,we get $\alpha = 17$,$\beta = -16$,$\gamma = 17$,and $\delta = 225$. Now,calculate $(\alpha + \beta + \gamma - \sqrt{\delta})^2$: $(17 - 16 + 17 - \sqrt{225})^2 = (18 - 15)^2 = 3^2 = 9$.

When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi}{4}$,if the equation $25x^2+9y^2=225$ is transformed to $\alpha x^2+\beta xy+\gamma y^2=\delta$,then $(\alpha+\beta+\gamma-\sqrt{\delta})^2=$

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