If the axes are rotated through an angle 45^{ | vedclass

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(A) Given the rotation angle $	heta = 45^{\circ}$.Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.The transformation equati

Vedclass · Accepted Answer

$(x+y)^2=4\sqrt{2}a(x-y)$

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(A) Given the rotation angle $	heta = 45^{\circ}$.Let the new coordinates be $(x', y')$ and the old coordinates be $(x, y)$.The transformation equations are:$x = x' \cos 45^{\circ} - y' \sin 45^{\circ} = \frac{x'-y'}{\sqrt{2}}$$y = x' \sin 45^{\circ} + y' \cos 45^{\circ} = \frac{x'+y'}{\sqrt{2}}$Substituting these into the equation $y^2 = 4ax$:$(\frac{x'+y'}{\sqrt{2}})^2 = 4a(\frac{x'-y'}{\sqrt{2}})$$\frac{(x'+y')^2}{2} = \frac{4a(x'-y')}{\sqrt{2}}$$(x'+y')^2 = \frac{8a}{\sqrt{2}}(x'-y') = 4\sqrt{2}a(x'-y')$Replacing $(x', y')$ with $(x, y)$,the transformed equation is $(x+y)^2 = 4\sqrt{2}a(x-y)$.

If the axes are rotated through an angle $45^{\circ}$ about the origin in an anticlockwise direction,then the transformed equation of $y^2=4ax$ is

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