If the axes are rotated through an angle $45^{\circ}$ about the origin in anticlockwise direction, then the…
If the axes are rotated through an angle $45^{\circ}$ about the origin in anticlockwise direction, then the transformed equation of $y^2=4 a x$ is
$(x+y)^2=4 \sqrt{2} a(x-y)$
$(x-y)^2=4 \sqrt{2} a(x+y)$
$(x-y)^2=\frac{4 a}{\sqrt{2}}(x+y)$
$(x+y)^2=\frac{4 a}{\sqrt{2}}(x-y)$
Solution
Given $\theta=45^{\circ}$
Let new co-ordinate be $\left(x^{\prime}, y^{\prime}\right)$ and old coordinate be $(x, y)$
So, $x=x^{\prime} \cos 45^{\circ}-y^{\prime} \sin 45^{\circ}=\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}$ and $y=x^{\prime} \sin 45^{\circ}+y^{\prime} \cos 45^{\circ}=\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}$ Now, $y^2=4 \mathrm{a} x$
$\begin{aligned} & \text { Now, } y^2=4 \mathrm{ax} \\ & \Rightarrow\left(\frac{x^{\prime}+y^{\prime}}{\sqrt{2}}\right)^2=4 a\left(\frac{x^{\prime}-y^{\prime}}{\sqrt{2}}\right) \Rightarrow\left(x^{\prime}+y^{\prime}\right)^2 \\ & =4 a \sqrt{2}\left(x^{\prime}-y^{\prime}\right)\end{aligned}$
So, $(x+y)^2=4 \sqrt{2} a(x-y)$