Transformation of Rectangular Axes - A Detailed Note
Published by Ganit Charcha
| Category - High School Math Topics | 2025-09-17 07:49:54
In coordinate geometry, transforming the axes allows us to simplify equations, reposition curves, and reveal hidden symmetries. This technique is especially powerful when dealing with conic sections and quadratic forms. Transformation of rectangular axes changes a point's coordinates by shifting or rotating the coordinate system itself. The two primary types are translation, which moves the origin without changing the axes' direction, and rotation, which pivots the axes around the origin. The transformation allows for easier analysis of geometric shapes by simplifying their equations in a new coordinate system. 1. Translation of Axes Definition: Shifting the origin from $(0, 0)$ to a new point $(h, k)$. This creates a new set of axes $(x’, y’)$ that are parallel to the original axes $(x, y)$. This help remove constant terms from an equation, simplifying it when the curve is not centered at the origin. Transformation Equations: $x = x' + h$ and $y = y' + k$ Geometric Interpretation: The entire coordinate grid slides without rotation. Useful for centering a curve at a more convenient point. Effect on Equations: Linear terms in a quadratic equation can often be eliminated. Example: Transforming $x^2+ y^2 + 4x - 6y + 9 = 0$ by shifting origin to $(-2, 3)$ simplifies the equation to a standard circle.
2. Rotation of Axes Definition: Rotating the coordinate axes by an angle (\theta), keeping the origin fixed. This helps to eliminate the xy-term from a quadratic equation, which simplifies the form of conic sections like ellipses or hyperbolas. Transformation Equations: $x = x'\cos\theta - y'\sin\theta$ and $y = x'\sin\theta + y'\cos\theta$ Geometric Interpretation: 1. The axes pivot around the origin. 2. Points retain their relative positions but are expressed in a rotated frame. Effect on Equations: 1. Eliminates the cross-product term $xy$ in conic sections. 2. Example: The general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ becomes simpler if $B \ne 0$ and we rotate axes to eliminate $xy$. Angle of Rotation: To eliminate the $xy$ term, the rotation that is needed is give by: [$\tan(2\theta) = \frac{B}{A - C}$
3. Combined Transformation Definition: Applying both translation and rotation to reposition and reorient the coordinate system. Use Case: 1. Transforming a conic to its canonical form. 2. Revealing symmetry or simplifying integration limits. Steps: 1. Translate origin to center of the conic. 2. Rotate axes to eliminate (xy) term. 3. Rewrite the equation in new coordinates. Conceptual Insight Transforming axes is not just algebraic—it’s geometric storytelling. You're changing the lens through which the curve is viewed, often revealing its true nature. Translation repositions the scene. Rotation reorients the perspective. Together, they simplify complexity into clarity.
Worked Out Problems
Problem 1. Using the transformation $x = x’ + 2$ and $y = y’ - 1$, find out the transformed equation of the curve $9x^2 + 16 y^2 – 36x + 36 y – 96 = 0$. Solution: The given equation $9x^2 + 16 y^2 – 36x + 36 y – 96 = 0$, after transformation becomes
Problem 2. Using the transformation $x = x’cos 45 – y’sin 45 = \frac{1}{\sqrt(2)}(x’ – y’)$ and $y = x’sin 45 + y’ cos 45 = \frac{1}{\sqrt(2)}(x’ + y’)$, where $(x’, y’)$ represents the co-ordinates of the points corresponding to the new axes. Find out the transformed equation w.r.t the new axes of the equation $5x^2 – 4xy + 5y^2 = 21$. Solution: The given equation $5x^2 – 4xy + 5y^2 – 21$ after transformation becomes,
Problem 3. After applying a transformation the equation $x^2 + y^2 -4x – 2y + 3 = 0$ becomes $x’^2 + y’^2 + 2ax’ +c = 0$ and the equation $x^2 + y^2 -6x – 2y + 5 = 0$ becomes $x’2 + y’^2 + 2bx’ + c = 0$ in the new co-ordinate system. Find the new origin. Solution: Let, the new origin be $(\alpha, \beta)$. Therefore, we have $x = x’ + \alpha$ and $y = y’ + \beta$. After transformation, the equation $x^2 + y^2 -4x – 2y + 3 = 0$ becomes, $(x'+\alpha)^2 + (y' + \beta)^2 -4(x'+\alpha) – 2(y' + \beta) + 3 = 0$ which on simplification gives $x'^2 + y'^2 + 2(\alpha - 2)x' + 2(\beta - 1)y' + (3 - 4\alpha -2\beta + \alpha^2 + \beta^2 = 0$ and since this is equivalent to $x’^2 + y’^2 + 2ax’ +c = 0$, so we have $\alpha - 2 = a$, $\beta - 1 = 0$ and $c = 3 - 4\alpha -2\beta + \alpha^2 + \beta^2$
Again, the equation $x^2 + y^2 -6x – 2y + 5 = 0$ after transformation becomes, $(x'+\alpha)^2 + (y' + \beta)^2 -6(x'+\alpha) – 2(y' + \beta) + 5 = 0$ which on simplification gives $x'^2 + y'^2 + 2(\alpha - 3)x' + 2(\beta - 1)y' + (5 - 6\alpha -2\beta + \alpha^2 + \beta^2) = 0$ and since this equivalent to $x’^2 + y’^2 + 2bx’ +c = 0$, so we have $\alpha - 3 = b$, $\beta - 1 = 0$ and $c = 5 - 6\alpha -2\beta + \alpha^2 + \beta^2$.
The identity, $c = 3 - 4\alpha -2\beta + \alpha^2 + \beta^2$ yields $c = (\alpha - 2)^2 + (\beta - 1) ^2 + 3 - 5$ $\Rightarrow$ $a^2 = c + 2$ Similarly, from $c = 5 - 6\alpha -2\beta + \alpha^2 + \beta^2$, we get on simplification $b^2 = c + 5$. We have $b - a = (\alpha - 3) - (\alpha -2) = -1$, and $b^2 - a^2 = 3$ $\Rightarrow$ $b + a = -3$. Solving for and b, we get $a = -1$ and $b = -2$ and $c = a^2 - 2 = -1$.
Therefore, $\alpha = a + 2 = 1$ and $\beta = 1$, making $(1, 1)$ as the new origin.
Problem 4. After applying a translational transformation the equation $x^2 + y^2 – 8x – 5y + 7 = 0$ gets transformed to an equation which does not have any term involving $x’$ and $y’$. Find out the transformed equation w.r.t the new co-ordinate system $(x’, y’)$. Solution: Let, the new origin be $(\alpha, \beta)$. Therefore, we have $x = x’ + \alpha$ and $y = y’ + \beta$. After transformation, the equation $x^2 + y^2 - 8x – 5y + 7 = 0$ becomes, $(x'+\alpha)^2 + (y' + \beta)^2 - 8(x'+\alpha) – 5(y' + \beta) + 7 = 0$ which on simplification yields $x'^2 + y'^2 + (2\alpha - 8)x' + (2\beta - 5)y' + (7 - 8\alpha - 5\beta + \alpha^2 + \beta^2) = 0$.
Since, transformed equation does not have terms involving $x'$ and $y'$, therefore, $2\alpha - 8 = 0$ which gives $\alpha = 4$ and $2\beta - 5 = 0$ which gives $\beta = \frac)5)(2)$. Hence, origin is shifted to $(4, \frac{5}{2})$, and the transformed equation becomes $x'^2 + y'^2 = \frac{61}{4}$.
Problem 5. By what angle the axes are required to be rotated so that the transformed equation of $7x^2 + 4xy + 3y^2 = 0$ will not have any term involving $x’y’$ w.r.t the new co-ordinate system $(x’, y’)$. Solution: Let, the axes be turned by the angle $\theta$. Hence the transformation equation is given by, $x = x'\cos\theta - y'\sin\theta$ and $y = x'\sin\theta + y'\cos\theta$.
On applying the transformation, the equation $7x^2 + 4xy + 3y^2 = 0$ becomes, $7x'^2 cos^2 \theta + 7y'^2 sin^2 \theta + 4(x'^2 - y'^2)sin \theta cos \theta + 3x'^2 sin^2 \theta + 3y'^2 cos^2 \theta -8 x'y' sin \theta cos \theta + 4x'y'(cos^2 \theta - sin^2 \theta)$
In the transformed equation, there is no term containg $x'y'$, so we have $-8 sin \theta cos \theta + 4(cos^2 \theta - sin^2 \theta) = 0$ $\Rightarrow$ $-4sin 2\theta + 4 cos 2\theta = 0$ $\Rightarrow$ $tan 2\theta = 1 = tan \frac{\pi}{4}$ $\Rightarrow$ $\theta = \frac{\pi}{8}$.
Problem 6. Keeping the origin fixed, if the axes are rotated through an ange $\theta$, the co-ordinate systems gets changed from $(x, y)$ to $(X, Y)$. If by applying the rotation, the equation $ux + vy = 0$ changes to $UX + VY = 0$, then prove that $u^2 + v^2 = U^2 + V^2.$ Solution: Let, the axes be turned by an angle $\theta$, keeping the origin fixed. Hence the transformation equation is given by, $x = X cos\theta - Y sin\theta$ and $y = X sin\theta + Y cos\theta$.
Now, $ux + vy = 0$ becomes after transformation, $u \, (X \, cos\theta - Y \, sin\theta) + v \, (X \, sin\theta + Y \, cos\theta) = 0$ $\Rightarrow (u \,cos \theta + v \,sin \theta)X + (v \, cos \theta - u \, sin \theta)Y = 0$
But, by question $ux + vy = 0$ becomes $UX + VY = 0$ and hence we have, $U = u \, cos \theta + v \, sin \theta$ and $V = v \, cos \theta - u \, sin \theta$
