If $f(x,y) =0 $ is an equation w.r.t $X,Y$ axes , the equation will be transformed to $f(x'cos \theta - y'sin \theta , x'sin \theta + y'cos \theta) = 0$ when the axes are transformed to an another axes of same origin. Let $X , Y$ axes are turned about $O$ through angle $\theta$ anti clockwisely to $X' ,Y'$.
Is the above statement correct?
If it is correct , I applied in the following..
If I want to transform $x^2 = 4y$ to $y^2 = 4x$ , we have to rotate the axes about the origin through $270$ degree anticlockwisely..So the new equation will be $(x'cos 270 - y'sin 270)^2 = 4(x'sin 270 + y'cos 270)$..But this way of transformation gives ${y'}^2 = -4x'$..
Can anyone please tell me where I am mistaking?
You’re making a very common error. In order to transform the first parabola into the second, you must rotate the parabola 90 degrees clockwise. This is equivalent to a counterclockwise rotation of the axes by the same amount.
Visualize the transformation of the axes as follows: The transformed parabola opens in the direction of the positive $x$-axis, so the new positive $x$-axis must be in the direction of the original positive $y$-axis. Since the transformation is a rigid motion, this also means that the the positive direction of new $y$-axis must correspond to the original negative $x$-axis.
Algebraically, if the new coordinates are related to the original ones via some invertible transformation $(x',y')=\varphi(x,y)$, in order to obtain the transformed equation you have to substitute for $x$ and $y$, but that means that you need the inverse transformation $(x,y)=\varphi^{-1}(x',y')$. Thus, for your example, to rotate the graph clockwise you have to rotate the axes counterclockwise. This is the same principle that you’ve already seen when translating and scaling graphs: to shift the graph to the right (positive shift) you have to subtract the distance to be shifted from $x$ and to scale by a factor of $\gamma$ in the $x$-direction you divide by $\gamma$.