Mathematics.

conic sections

Translations of Conics

Analytic Geometry25 minDifficulty5 out of 10

You should know: conic sections

Overview

The standard-form equations for conics (centered at the origin, vertex at the origin) are the special case where the curve happens to sit at the coordinate system's center. Sliding — translating — a conic to a new center (h, k) is done by the same substitution that shifts any graph: replace x with (x − h) and y with (y − k) everywhere in the standard equation. This turns the clean standard forms into the general translated forms seen throughout applications, and running the substitution in reverse — completing the square on a messy second-degree equation — recovers the center, axes, and eccentricity of a conic given in expanded form.

Intuition

Translating a conic is exactly like sliding a stencil across a page: the shape — its size, eccentricity, orientation — never changes, only its position does. Algebraically this is the same x → x−h, y → y−k substitution used to shift any graph, because subtracting h from every x-value in the equation shifts the whole picture right by h (and similarly for k). Running this backward is the useful direction in practice: a general second-degree equation with linear terms (like x² − 6x + y² + 4y − 3 = 0) is really a standard conic in disguise, and completing the square on the x-terms and y-terms separately un-shifts it back to reveal the center and standard form.

Formal Definition

Definition

Translating each standard conic so its center (or vertex, for a parabola) moves from the origin to the point (h, k):

(x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

Center (h,k); foci at (h\pm c, k) if the major axis is horizontal, c^2=a^2-b^2

Translated ellipse
(x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1

Center (h,k); foci at (h\pm c,k), c^2=a^2+b^2

Translated hyperbola (horizontal transverse axis)
(y−k)2=4p(x−h)(y-k)^2 = 4p(x-h)

Vertex (h,k); focus (h+p, k); directrix x = h-p

Translated parabola (opens horizontally)
(x−h)2=4p(y−k)(x-h)^2 = 4p(y-k)

Vertex (h,k); focus (h, k+p); directrix y = k-p

Translated parabola (opens vertically)

Derivation

Completing the square converts a general (translated, axis-aligned) conic equation into standard form. For example, starting from a circle-like equation with linear terms:

x2+y2−6x+4y−3=0x^2 + y^2 - 6x + 4y - 3 = 0

Group x-terms and y-terms

(x2−6x)+(y2+4y)=3(x^2-6x) + (y^2+4y) = 3

Move the constant to the right

(x2−6x+9)+(y2+4y+4)=3+9+4(x^2-6x+9) + (y^2+4y+4) = 3+9+4

Add (half the linear coefficient)² to both sides for each variable

(x−3)2+(y+2)2=16(x-3)^2 + (y+2)^2 = 16

This is a circle of radius 4 centered at (3, -2)

Properties

Shape invariance

Translation preserves a,b,c,e exactly — only the center/vertex location (h,k) changes\text{Translation preserves } a, b, c, e \text{ exactly — only the center/vertex location } (h,k) \text{ changes}

Foci shift with center

Every focus and vertex shifts by the same (h,k) as the center/vertex of the conic\text{Every focus and vertex shifts by the same } (h,k) \text{ as the center/vertex of the conic}

Axes remain parallel to coordinate axes

Translation alone (no rotation) keeps the major/minor or transverse/conjugate axes parallel to the x- and y-axes\text{Translation alone (no rotation) keeps the major/minor or transverse/conjugate axes parallel to the } x\text{- and } y\text{-axes}

General second-degree form

Ax2+Cy2+Dx+Ey+F=0 (no xy term) always reduces to a translated conic (or degenerate case) via completing the squareAx^2+Cy^2+Dx+Ey+F=0 \ (\text{no } xy \text{ term}) \text{ always reduces to a translated conic (or degenerate case) via completing the square}

Applications

Satellite dish and headlight reflector designs are specified as parabolas with a vertex placed wherever the mounting geometry requires — a translated, not origin-centered, parabola.

Worked Examples

  1. Substitute h=2, k=-3, a=5, b=3 into the translated ellipse form.

    (x−2)225+(y+3)29=1\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1

  2. Compute c using c² = a² − b², then shift the foci by (h,k) = (2,-3).

    c2=25−9=16  ⟹  c=4  ⟹  F=(2±4, −3)c^2 = 25-9=16 \implies c=4 \implies F = (2\pm 4,\ -3)

Answer: (x−2)²/25 + (y+3)²/9 = 1, with foci at (6, −3) and (−2, −3).

Practice Problems

Difficulty 5/10

Find the center and foci of the hyperbola (x+1)²/9 − (y−2)²/16 = 1.

Difficulty 6/10

Complete the square to write x² + 4y² − 4x + 24y + 24 = 0 in standard translated-ellipse form, and find its center.

Difficulty 6/10

A satellite dish's cross-section is a parabola with vertex at ground-level point (0, 0.5) meters (0.5 m above the mounting base) opening upward, and the receiver (focus) sits 1.2 m above the vertex. Write the parabola's equation and find the directrix.

Common Mistakes

Common Mistake

Forgetting to shift the foci and vertices along with the center when translating, and leaving them at their origin-centered coordinates.

Every distinguished point of the conic — center, vertices, foci — must be shifted by the same (h, k); only the shape-determining lengths a, b, c stay fixed under translation.

Common Mistake

When completing the square, forgetting to multiply the added constant by the leading coefficient before moving it to the other side (e.g. in x² + 4y² − 4x + 24y + 24 = 0, adding 9 instead of 4·9=36 for the y-term).

Because the y² term has coefficient 4, completing the square inside 4(y²+6y) requires adding 4·9=36 to balance the equation, not just 9 — the coefficient must be factored out first or accounted for when balancing.

Quiz

Translating a conic from the origin to center (h, k) is achieved algebraically by replacing:
Translating an ellipse to a new center changes:
To find the center of a conic given as a general equation like x²+4y²−4x+24y+24=0, the standard technique is to:

Summary

  • Translating a conic moves its center/vertex from the origin to (h,k) via x→x−h, y→y−k, without changing its shape, size, or eccentricity.
  • Standard translated forms: (x−h)²/a²+(y−k)²/b²=1 (ellipse), (x−h)²/a²−(y−k)²/b²=1 (hyperbola), (y−k)²=4p(x−h) or (x−h)²=4p(y−k) (parabola).
  • Foci and vertices translate along with the center by the same (h,k); axis lengths a, b, c and eccentricity e are unaffected.
  • A general second-degree equation with no xy-term (Ax²+Cy²+Dx+Ey+F=0) can always be reduced to a translated standard form by completing the square in x and y separately.

References

  1. WebsiteWikipedia — Conic section
  2. WebsiteWikipedia — Ellipse