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# Translation of axes

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### Translation of axes

In mathematics, a translation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the x' axis is parallel to the x axis and k units away, and the y' axis is parallel to the y axis and h units away. This means that the origin O' of the new coordinate system has coordinates (h, k) in the original system. The positive x' and y' directions are taken to be the same as the positive x and y directions. A point P has coordinates (x, y) with respect to the original system and coordinates (x', y') with respect to the new system, where

x = x' + h      and      y = y' + k

(1)

or equivalently

x' = x - h      and      y' = y - k .[1][2]

(2)

In the new coordinate system, the point P will appear to have been translated in the opposite direction. For example, if the xy-system is translated a distance h to the right and a distance k upward, then P will appear to have been translated a distance h to the left and a distance k downward in the x'y'-system . A translation of axes in more than two dimensions is defined similarly.[3] A translation of axes is a rigid transformation, but not a linear map. (See Affine transformation.)

## Contents

• Motivation 1
• Translation of conic sections 2
• Example 1 2.1
• Generalization to several dimensions 3
• Translation of quadric surfaces 4
• Example 2 4.1
• Notes 5
• References 6

## Motivation

When we want to study the equations of curves and when we wish to use the methods of analytic geometry, coordinate systems become essential. When we use the method of coordinate geometry we place the axes at a position that is "convenient" with respect to the curve under consideration. For example, when we study the equations of ellipses and hyperbolas, the foci are usually located on one of the axes and are situated symmetrically with respect to the origin. But now suppose that we have a problem in which the curve (hyperbola, parabola, ellipse, etc.) is not situated so conveniently with respect to the axes. We would then like to change the coordinate system in order to have the curve at a convenient and familiar location and orientation. The process of making this change is called a transformation of coordinates.[4]

The solutions to many problems can be simplified by translating the coordinate axes to obtain new axes parallel to the original ones.[5]

## Translation of conic sections

Through a change of coordinates, the equation of a conic section can be put into a standard form, which is usually easier to work with. For the most general equation of the second degree, it is always possible to perform a rotation of axes in such a way that in the new system the equation takes the form

Ax^2 + Cy^2 + Dx + Ey + F = 0      (A and C not both zero),

(3)

that is, there is no xy term.[6] We assume here that this has already been done. We now illustrate how to use a translation of axes to reduce an equation of the form (3) to an equation of the same form but with new variables (x', y') as coordinates, and with D and E both equal to zero (with certain exceptions – for example, parabolas). The principal tool in this process is "completing the square."[7]

### Example 1

Given the equation

9x^2 + 25y^2 + 18x - 100y - 116 = 0 ,

by using a translation of axes determine whether the locus of the equation is a parabola, ellipse, or hyperbola. Determine foci (or focus), vertices (or vertex), and eccentricity.

Solution: To complete the square in x and y, we write the equation in the form

9(x^2 + 2x \qquad ) + 25(y^2 - 4y \qquad ) = 116 .

We complete the squares and obtain

9(x^2 + 2x + 1) + 25(y^2 - 4y + 4) = 116 + 9 + 100
\Leftrightarrow 9(x + 1)^2 + 25(y - 2)^2 = 225 .

We define

x' = x + 1      and      y' = y - 2 .

That is, the translation in equations (2) is made with h = -1, k = 2 . The equation in the new coordinate system is

9x'^2 + 25y'^2 = 225 .

Dividing by 225, we find

\frac{x'^2}{25} + \frac{y'^2}{9} = 1 ,

which we recognize as an ellipse with a = 5, b = 3, c^2 = a^2 - b^2 = 16, c = 4, e = \tfrac{4}{5} . In the x'y'-system, we have: center (0, 0) ; vertices (\pm 5, 0) ; foci (\pm 4, 0) .

In the xy-system, we use the relations x = x' - 1, y = y' + 2 to obtain: center (-1, 2) ; vertices (4, 2), (-6, 2) ; foci (3, 2), (-5, 2) ; eccentricity \tfrac{4}{5} .[8]

## Generalization to several dimensions

Suppose that we have an xyz-Cartesian coordinate system in three dimensions and we introduce a second Cartesian coordinate system, with axes x', y' and z' so located that the x' axis is parallel to the x axis and h units from it, the y' axis is parallel to the y axis and k units from it, and the z' axis is parallel to the z axis and l units from it. A point P in space will have coordinates in both systems. If its coordinates are (x, y, z) in the original system and (x', y', z') in the second system, the equations

x' = x - h, \qquad y' = y - k, \qquad z' = z - l

(4)

hold.[9] Equations (4) define a translation of axes in three dimensions where (h, k, l) are the xyz-coordinates of the new origin.[10] A translation of axes in any finite number of dimensions is defined similarly.

In three-space, the most general equation of the second degree in x, y and z has the form

Ax^2 + By^2 + Cz^2 + Dxy + Exz + Fyz + Gx + Hy + Kz + L = 0 ,

(5)

where the quantities A, B, C, \ldots , L are positive or negative numbers or zero. The points in space satisfying such an equation all lie on a surface. Any second-degree equation which does not reduce to a cylinder, plane, line, or point corresponds to a surface which we call quadric.[11]

As in the case of plane analytic geometry, the method of translation of axes may be used to simplify second-degree equations, thereby making evident the nature of certain quadric surfaces. The principal tool in this process is "completing the square."[12]

### Example 2

Use a translation of coordinates to identify the quadric surface

x^2 + 4y^2 + 3z^2 + 2x - 8y + 9z = 10 .

Solution: We write

x^2 + 2x \qquad + 4(y^2 - 2y \qquad ) + 3(z^2 + 3z \qquad ) = 10 .

Completing the square, we obtain

(x + 1)^2 + 4(y - 1)^2 + 3(z + \tfrac{3}{2})^2 = 10 + 1 + 4 + \tfrac{27}{4} .

Introducing the translation of coordinates

x' = x + 1, \qquad y' = y - 1, \qquad z' = z + \tfrac{3}{2} ,

we find for the equation of the surface,

x'^2 + 4y'^2 + 3z'^2 = \tfrac{87}{4} ,

which we recognize as the equation of an ellipsoid.[13]

## Notes

1. ^ Anton (1987, p. 107)
2. ^ Protter & Morrey (1970, p. 315)
3. ^ Protter & Morrey (1970, pp. 585-588)
4. ^ Protter & Morrey (1970, pp. 314-315)
5. ^ Anton (1987, p. 107)
6. ^ Protter & Morrey (1970, p. 322)
7. ^ Protter & Morrey (1970, p. 316)
8. ^ Protter & Morrey (1970, pp. 316-317)
9. ^ Protter & Morrey (1970, pp. 585-586)
10. ^ Anton (1987, p. 107)
11. ^ Protter & Morrey (1970, p. 579)
12. ^ Protter & Morrey (1970, p. 586)
13. ^ Protter & Morrey (1970, p. 586)

## References

• Anton, Howard (1987), Elementary Linear Algebra (5th ed.), New York:
• Protter, Murray H.; Morrey, Jr., Charles B. (1970), College Calculus with Analytic Geometry (2nd ed.), Reading:
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