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

Vedclass · Accepted Answer

$1225$

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(B) The transformation equations for rotation of axes by an angle $	heta = \frac{\pi}{4}$ 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 $ax^2+2hxy+by^2=c$: $a(\frac{X-Y}{\sqrt{2}})^2 + 2h(\frac{X-Y}{\sqrt{2}})(\frac{X+Y}{\sqrt{2}}) + b(\frac{X+Y}{\sqrt{2}})^2 = c$ $\frac{a}{2}(X^2-2XY+Y^2) + h(X^2-Y^2) + \frac{b}{2}(X^2+2XY+Y^2) = c$ $X^2(\frac{a+2h+b}{2}) + XY(b-a) + Y^2(\frac{a-2h+b}{2}) = c$ Comparing with $25X^2+9Y^2=225$,we get: $\frac{a+2h+b}{2} = 25 \implies a+2h+b = 50$ $b-a = 0 \implies a=b$ $\frac{a-2h+b}{2} = 9 \implies a-2h+b = 18$ $c = 225 \implies \sqrt{c} = 15$ Thus,$(a+2h+b-\sqrt{c})^2 = (50-15)^2 = 35^2 = 1225$.

When the coordinate axes are rotated about the origin through an angle $\frac{\pi}{4}$ in the positive direction,the equation $ax^2+2hxy+by^2=c$ is transformed to $25x^2+9y^2=225$,then $(a+2h+b-\sqrt{c})^2=$

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