Transformations of Axes

In this article, we will cover the concept of Transformation of axes. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of six questions have been asked on JEE MAINS( 2013 to 2023) from this topic in the last few years.

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This Story also Contains

1. What are the Transformations of Axes?
2. Types of Transformation of Axes
3. Shifting of Origin: Definition
4. Procedure for calculating new coordinates
5. Solved Examples Based on the Transformation of Axes

What are the Transformations of Axes?

To use the method of coordinate geometry, the axes are placed at a convenient position with respect to the curve under consideration. If the curve (parabola, ellipse, hyperbola, etc.) is not situated conveniently with respect to the axes, the coordinate system should be changed to place the curve at a convenient and familiar location and orientation. The process of making this change is called the transformation of coordinates.

Types of Transformation of Axes

There are basically two types of transformations:

1) Shifting of origin (translation of axes)

2) Rotation of axes.

3) Dilation

4) Reflection

Shifting of Origin: Definition

The shifting (translation) of the coordinate axes is done without rotation so that each axis remains parallel to its original position. In a translation, each point of the shape must be moved in the same direction and for the same distance. When you are doing a translation, the primary object is called the pre-image, and the object after the translation is called the image.

Procedure for calculating new coordinates

Point P has coordinate $(\mathrm{x},\mathrm{y})$ in the original coordinate system, i.e. in the $xy-$coordinate system.

If the new origin takes a position as $O^{\prime }(h,k)$ with new$x$ and $y$ axes remaining parallel to the old axes.

The coordinates of the point P are now $(X,Y)=(x-h,y-k)$ w.r.t. the new coordinate system (i.e. $Y^{\prime }{O}^{\prime }{X}^{\prime }$).

Thus, $\mathrm{X}=\mathrm{x}-\mathrm{h}$ and $Y=y-k$

Or, $x=X+h$ and $y=Y+k$

Note:

If the function $f(x,y)=0$ is with respect to the original coordinate system, then the equation with respect to the new coordinate system is $f(x+h,y+k)$ $=0$.

Rotation of Axes About Origin

Another type of transformation of axes is the rotation of axes. The rotation of Axes about the origin is done without shifting the axes.

Recommended Video Based on the Transformation of Axes

Solved Examples Based on the Transformation of Axes

Example 1: What are the coordinates of $P(2,3)$ when the origin is shifted from $(0,0)$ to $(4,5)$?

Solution: If the origin is shifted from $(0,0)$ to $(h,k)$; then the new coordinates for $(x,y)$ are $(x-h,y-k)$.

So, new coordinates for P are

$\begin{array}{rl}& (2-4,3-5)\\ & =(-2,-2)\end{array}$

Hence, the answer is $(-2,-2)$.

Example 2: At what point should the origin be shifted so that the coordination of points $(2,1)$ become $(3,4)$?

Solution: Let the origin be shifted to $(h,k)$

The new coordinates are ( 2-h, 1-k)

Given this equals $(3,4)$

So, $2-h=3$, hence $h=-1$.

and $1-k=4$, hence $k=-3$.

So, the origin should be shifted to $(-1,-3)$.

Hence, the answer is $(-1,-3)$.

Example 3: Without changing the direction of coordinates axes, the origin is transferred to $(\alpha ,\beta )$ so that the linear terms in the equation $x^2+y^2+2x-4y+6=0$ are eliminated. The point $(\alpha ,\beta )$ is

Solution: The given equation is $x^2+y^2+2x-4y+6=0\dots ..(1)$

Putting $\mathrm{x}={\mathrm{x}}^{\prime }+\alpha$ and $\mathrm{y}={\mathrm{y}}^{\prime }+\beta \mathrm{in}(1)$ we ${\mathrm{getx}}^{\prime 2}+{\mathrm{y}}^{\prime 2}+{\mathrm{x}}^{\prime }(2\alpha +2)+{\mathrm{y}}^{\prime }(2\beta -4)+({\alpha }^2+{\beta }^2+2\alpha -4\beta +6)=0$.
To eliminate linear terms, we should have $2\alpha +2=0$ and $2\beta -4=0$.
$\begin{array}{rl}& \Rightarrow \alpha =-1\text{ and }\beta =2\\ & \Rightarrow (\alpha ,\beta )\equiv (-1,2)\end{array}$
Hence, the answer is $(-1,2)$.

Example 4: Point $\mathrm{P}(\mathrm{a},\mathrm{b})$ lie on the line $y=x+1.P$ is shifted in the direction perpendicular to the given line so that it meets $\mathrm{X}$-axis at $(\alpha ,0)$, then

Solution: The parametric equation of the line passing through $(a,b)$ and $\perp$ to $y=x+1$ is

$\frac{\mathrm{x}-\mathrm{a}}{-\frac{1}{\sqrt{2}}}=\frac{\frac{\mathrm{y}-\mathrm{b}}{1}}{\sqrt{2}}=\alpha$
Any point on this line is given by $(a+\frac{\alpha }{\sqrt{2}},b+\frac{\alpha }{\sqrt{2}})$.
If this point is lying on the X-axis, then $\mathrm{b}+\frac{\frac{\alpha }{\sqrt{2}}}{1}=0\Rightarrow \alpha =-\sqrt{2}\mathrm{b}$
Hence, the point is $(a-b,0)$.

Hence, the answer is $\alpha =2a+1$.