Coordinate System and Coordinates
17.0 Rotation of axes
17.0 Rotation of axes
To find the new coordinates of a point when the direction of axes is rotated through an angle $\theta $.
Condition: The origin $O$ must be fixed.
Procedure: As shown in figure,
- Let $OX$ and $OY$ be the original system of co-ordinate axes and $OX'$ and $OY'$ be the new axes obtained after rotating the original axes through an angle $\theta $. Let $P(x,y)$ be a point of reference to the original axes and $P(X, Y)$ be a point with reference to the new axes.
- From figure, $OM=x$, $PM=y$, $ON=X$ and $PN=Y$ and we can write $$\begin{equation} \begin{aligned} OM = OL - ML \\ OM = OL - QN \\ OM = ON\cos \theta - PN\sin \theta \\ x = X\cos \theta - Y\sin \theta ...(1) \\\end{aligned} \end{equation} $$ Similarly, $$\begin{equation} \begin{aligned} PM = PQ + QM \\ PM = PQ + NL \\ PM = PN\cos \theta + ON\sin \theta \\ y = Y\cos \theta + X\sin \theta \\ y = X\sin \theta + Y\cos \theta ...(2) \\\end{aligned} \end{equation} $$
Now, $$(1) \times cos\theta + (2) \times \sin \theta $$$$X = x\cos \theta + y\sin \theta ...(3)$$ and $$(2) \times \cos \theta - (1) \times \sin \theta $$$$Y = - x\sin \theta + y\cos \theta ...(4)$$
Result:
- When the axes are rotated through an angle $\theta $, replace $(x,y)$ by $(x\cos \theta - y\sin \theta ,x\sin \theta + y\cos \theta )$ to find the co-ordinates or equation of curve with respect to new axes.
- When the co-odinates or equation of curve with respect to new axes are given and we have to find out the original co-ordinates or equation of curve, replace $(x,y)$ by $(x\cos \theta + y\sin \theta , - x\sin \theta + y\cos \theta )$.
Note: $${x^2} + {y^2} = {X^2} + {Y^2}$$ i.e., the distance of point $P$ from origin $O$ remains same even after the rotation of axes.
Question 8. If the axes are turned through ${45^ \circ }$, find the transformed form of the equation $3{x^2} + 3{y^2} + 2xy = 2$.
Solution: As given in the question $\theta = {45^ \circ }$, $\sin \theta = \cos \theta = \frac{1}{{\sqrt 2 }}$
Replacing $(x,y)$ by $(x\cos \theta - y\sin \theta ,x\sin \theta + y\cos \theta )$ i.e., $(\frac{{x - y}}{{\sqrt 2 }},\frac{{x + y}}{{\sqrt 2 }})$
The transformed form of the equation becomes $$\begin{equation} \begin{aligned} 3{(\frac{{x - y}}{{\sqrt 2 }})^2} + 3{(\frac{{x + y}}{{\sqrt 2 }})^2} + 2(\frac{{x - y}}{{\sqrt 2 }})(\frac{{x + y}}{{\sqrt 2 }}) = 2 \\ 3(2{x^2} + 2{y^2}) + 2({x^2} - {y^2}) = 4 \\ 8{x^2} + 4{y^2} = 4 \\ 2{x^2} + {y^2} = 1 \\\end{aligned} \end{equation} $$
