When the coordinate axes are rotated through | vedclass

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Question

(A) Let the original coordinates be $(X, Y)$ and the new coordinates be $(x, y)$.Given the rotation angle $	heta = \frac{\pi}{4}$,the transformation

Vedclass · Accepted Answer

$x^2+y^2-7\sqrt{2}x+\sqrt{2}y+21=0$

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(A) 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	heta - y \sin	heta = x \frac{1}{\sqrt{2}} - y \frac{1}{\sqrt{2}} = \frac{x-y}{\sqrt{2}}$$Y = x \sin	heta + y \cos	heta = x \frac{1}{\sqrt{2}} + y \frac{1}{\sqrt{2}} = \frac{x+y}{\sqrt{2}}$Substituting these into the transformed equation $x^2+y^2-6x+8y+21=0$ is incorrect because the question asks for the original equation given the transformed one.Actually,if the original equation is $f(X, Y) = 0$,the transformed equation is $f(x \cos	heta - y \sin	heta, x \sin	heta + y \cos	heta) = 0$.Given $x^2+y^2-6x+8y+21=0$ is the transformed equation,we substitute $x = X \cos	heta + Y \sin	heta$ and $y = -X \sin	heta + Y \cos	heta$ where $	heta = \frac{\pi}{4}$.$x = \frac{X+Y}{\sqrt{2}}, y = \frac{-X+Y}{\sqrt{2}}$Substituting into $x^2+y^2-6x+8y+21=0$:$(\frac{X+Y}{\sqrt{2}})^2 + (\frac{-X+Y}{\sqrt{2}})^2 - 6(\frac{X+Y}{\sqrt{2}}) + 8(\frac{-X+Y}{\sqrt{2}}) + 21 = 0$$\frac{X^2+Y^2+2XY}{2} + \frac{X^2+Y^2-2XY}{2} - \frac{6X+6Y}{\sqrt{2}} + \frac{-8X+8Y}{\sqrt{2}} + 21 = 0$$X^2+Y^2 - \frac{14X}{\sqrt{2}} + \frac{2Y}{\sqrt{2}} + 21 = 0$$X^2+Y^2 - 7\sqrt{2}X + \sqrt{2}Y + 21 = 0$.

When the coordinate axes are rotated through an angle $\frac{\pi}{4}$ in the positive direction,an equation is transformed to $x^2+y^2-6x+8y+21=0$. Then the original equation is

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