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(B) The transformation equations for rotation of axes by an angle $	heta = \frac{\pi}{4}$ are given by: $x = X \cos 	heta - Y \sin 	heta = \frac{X-

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The transformation equations for rotation of axes by an angle $	heta = \frac{\pi}{4}$ are given by: $x = X \cos 	heta - Y \sin 	heta = \frac{X-Y}{\sqrt{2}}$ $y = X \sin 	heta + Y \cos 	heta = \frac{X+Y}{\sqrt{2}}$ Substituting these into the transformed equation $X^2+Y^2-6X+8Y+21=0$: $\left(\frac{X-Y}{\sqrt{2}}ight)^2 + \left(\frac{X+Y}{\sqrt{2}}ight)^2 - 6\left(\frac{X-Y}{\sqrt{2}}ight) + 8\left(\frac{X+Y}{\sqrt{2}}ight) + 21 = 0$ $\frac{X^2+Y^2-2XY}{2} + \frac{X^2+Y^2+2XY}{2} - 3\sqrt{2}(X-Y) + 4\sqrt{2}(X+Y) + 21 = 0$ $X^2 + Y^2 - 3\sqrt{2}X + 3\sqrt{2}Y + 4\sqrt{2}X + 4\sqrt{2}Y + 21 = 0$ $X^2 + Y^2 + \sqrt{2}X + 7\sqrt{2}Y + 21 = 0$ Comparing with $ax^2+by^2+cx+dy+e=0$,we get $a=1, b=1, c=\sqrt{2}, d=7\sqrt{2}, e=21$. Now,calculate $(a+b+c^2+d^2-5e)^2$: $(1+1+(\sqrt{2})^2+(7\sqrt{2})^2-5(21))^2$ $= (2 + 2 + 98 - 105)^2$ $= (102 - 105)^2 = (-3)^2 = 9$.

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