Analytic Geometry — Parabolas, Ellipses, and Hyperbolas

Conic sections are curves described by distances. A parabola keeps equal distance from a focus and a line, an ellipse keeps the sum of distances to two foci constant, and a hyperbola keeps their difference constant. Their standard equations make it possible to find key points and sketch a reliable graph without plotting many random points. Familiarity with coordinate-plane distance and function transformations is helpful before beginning.

Recognising the Three Families

After an equation has been put into standard form, its squared terms reveal the curve:

Curve	Standard form	Key visual feature
Parabola	(x − h)² = 4p(y − k) or (y − k)² = 4p(x − h)	One squared variable
Ellipse	(x − h)²/a² + (y − k)²/b² = 1	Two positive squared terms
Hyperbola	(x − h)²/a² − (y − k)²/b² = 1, or the signs reversed	Two squared terms with opposite signs

In every form, (h, k) is the translated centre or vertex. Be alert: a term such as (x + 3)² means x − (−3), so its coordinate is −3.

Parabolas: Focus and Directrix

A parabola is the set of points equally far from a fixed point, the focus, and a fixed line, the directrix. The vertex is halfway between them. In the vertical form (x − h)² = 4p(y − k), the focus is (h, k + p) and the directrix is y = k − p. A positive p opens upward; a negative p opens downward.

For the horizontal form (y − k)² = 4p(x − h), the focus is (h + p, k) and the directrix is x = h − p. Its axis of symmetry is horizontal.

Worked example: read a translated parabola

Find the vertex, focus, directrix, and opening direction of (x − 2)² = −12(y + 1).

Compare with (x − h)² = 4p(y − k).
h = 2, k = −1, and 4p = −12, so p = −3.

Vertex:    (2, −1)
Focus:     (h, k + p) = (2, −4)
Directrix: y = k − p = −1 − (−3) = 2

Because p is negative, the parabola opens downward.
Its vertical axis of symmetry is x = 2.

Worked example: translate from vertex form

Write the standard focus-directrix equation for y = (1/8)(x + 1)² + 3.

y − 3 = (1/8)(x + 1)²
8(y − 3) = (x + 1)²

(x + 1)² = 8(y − 3) = 4p(y − 3)
4p = 8, so p = 2.

Vertex: (−1, 3)     Focus: (−1, 5)
Directrix: y = 1

Ellipses: A Fixed Total Distance

An ellipse consists of points for which the distances to two foci add to a constant. Its centre is (h, k). The larger denominator is a²; it lies under the coordinate in the direction of the major axis. The smaller denominator is b², for the minor axis. For an ellipse, a > b and

c² = a² − b².

The foci are c units from the centre along the major axis. If the larger denominator is under x, vertices are (h ± a, k) and foci are (h ± c, k). If it is under y, interchange the roles of x and y.

Worked example: sketch an ellipse from its equation

Analyse (x − 4)²/25 + (y + 2)²/9 = 1.

Centre: (h, k) = (4, −2)
25 is larger and is under x: the major axis is horizontal.
a = 5, b = 3
c² = a² − b² = 25 − 9 = 16, so c = 4.

Vertices:       (4 ± 5, −2) = (−1, −2), (9, −2)
Co-vertices:    (4, −2 ± 3) = (4, −5), (4, 1)
Foci:           (4 ± 4, −2) = (0, −2), (8, −2)

For a first sketch, plot the centre, vertices, and co-vertices; draw a smooth closed oval through the four axis endpoints. The foci lie inside, not on, the ellipse.

Worked example: build an ellipse equation

An ellipse has centre (−2, 1), horizontal major axis of length 10, and minor axis of length 6. Find its equation and foci.

Major-axis length = 2a = 10, so a = 5.
Minor-axis length = 2b = 6, so b = 3.

(x + 2)²/25 + (y − 1)²/9 = 1

c² = 25 − 9 = 16, so c = 4.
Foci: (−2 ± 4, 1) = (−6, 1), (2, 1)

Hyperbolas: A Fixed Difference

A hyperbola has two separate branches. For each point, the absolute difference between its distances to the foci is constant. In a horizontal hyperbola,

(x − h)²/a² − (y − k)²/b² = 1,

the vertices are (h ± a, k), the foci are (h ± c, k), and c² = a² + b². Notice the plus sign here, unlike the ellipse. The asymptotes, lines the branches approach without meeting, are

y − k = ±(b/a)(x − h).

When the positive fraction is the y-squared term, the hyperbola is vertical. Its vertices are (h, k ± a) and its asymptotes are y − k = ±(a/b)(x − h).

Worked example: find a vertical hyperbola's features

Analyse (y − 1)²/16 − (x + 3)²/9 = 1.

Centre: (−3, 1)
The positive term is the y-term, so the branches open up and down.
a = 4, b = 3
c² = a² + b² = 16 + 9 = 25, so c = 5.

Vertices: (−3, 1 ± 4) = (−3, −3), (−3, 5)
Foci:    (−3, 1 ± 5) = (−3, −4), (−3, 6)
Asymptotes:
y − 1 = ±(a/b)(x + 3) = ±(4/3)(x + 3)

To sketch, draw a light rectangle centred at (−3, 1) with half-width b = 3 and half-height a = 4. Its diagonals are the asymptotes. Start each branch at a vertex and curve outward toward both appropriate asymptotes.

Translating General Equations

Equations often arrive expanded. Group the x-terms and y-terms, complete the square in each group, then divide so the right side becomes 1. Completing the square means adding the square of half a linear coefficient inside a group and balancing that addition on the other side.

Worked example: complete the square for an ellipse

Put 9x² + 4y² − 36x + 16y − 56 = 0 into standard form.

9(x² − 4x) + 4(y² + 4y) = 56
9[(x − 2)² − 4] + 4[(y + 2)² − 4] = 56
9(x − 2)² + 4(y + 2)² = 108

(x − 2)²/12 + (y + 2)²/27 = 1

This is an ellipse centred at (2, −2), with a vertical major axis.

Applications and Meaning

Conics are not merely drawing exercises. A parabolic reflector sends rays parallel to its axis through its focus, which is why satellite dishes, headlights, and solar concentrators place a receiver or lamp there. Elliptical rooms and reflectors can direct sound or light from one focus toward the other. Hyperbolas appear in time-difference location systems: if a signal reaches two receivers at different times, the possible source positions form a hyperbola with the receivers as foci. In each application, the focus-distance definition explains the useful behaviour.

Common Mistakes to Avoid

Using the wrong sign for a translation: read (x + 4)² as (x − (−4))², so h = −4.
Forgetting the factor 4 in a parabola: in 4p, solve for p; do not call the entire coefficient the focal distance.
Choosing a by coordinate letter: for ellipses and hyperbolas, a² is determined by the major/transverse direction, not automatically by the x-denominator.
Subtracting for hyperbola foci: use c² = a² + b² for a hyperbola, but subtract for an ellipse.
Drawing hyperbola branches across the centre: the asymptotes cross at the centre, but the hyperbola does not.

Practice Set

For (x + 3)² = 20(y − 1), find the vertex, focus, directrix, axis, and opening direction.
Write the equation of a horizontal parabola with vertex (2, −1) and focus (−1, −1).
Find the centre, vertices, co-vertices, and foci of (x − 1)²/16 + (y + 4)²/7 = 1.
An ellipse is centred at (0, 3), has a vertical major axis, a = 6, and b = 2. Write its equation and locate its foci.
For (x + 2)²/4 − (y − 5)²/12 = 1, give the centre, vertices, foci, and asymptote equations.
Write an equation for a vertical hyperbola centred at (4, −2), with vertices (4, 1) and (4, −5), and conjugate-axis length 8.
Put x² + y² − 6x + 8y − 11 = 0 into standard form. Identify the special conic.
A parabolic dish has cross-section x² = 24y, with its vertex at the origin. How far from the vertex should a receiver be placed on the axis?

Answer Checks

4p = 20 gives p = 5. Vertex: (−3, 1); focus: (−3, 6); directrix: y = −4; axis: x = −3; it opens upward.
Here p = −3 because the focus is three units left of the vertex. Thus (y + 1)² = −12(x − 2).
Centre (1, −4); a = 4 and b = √7, so the major axis is horizontal. Vertices: (−3, −4), (5, −4). Co-vertices: (1, −4 ± √7). Since c² = 16 − 7 = 9, foci are (−2, −4), (4, −4).
x²/4 + (y − 3)²/36 = 1. Since c² = 36 − 4 = 32, the foci are (0, 3 ± 4√2).
Centre (−2, 5), a = 2, b = 2√3, and c = 4. Vertices: (−4, 5), (0, 5); foci: (−6, 5), (2, 5). Asymptotes: y − 5 = ±√3(x + 2).
The vertex distance is a = 3 and the conjugate-axis length is 2b = 8, so b = 4. The equation is (y + 2)²/9 − (x − 4)²/16 = 1.
x² − 6x + y² + 8y = 11 gives (x − 3)² + (y + 4)² = 36. It is a circle, the special ellipse with equal radii, centred at (3, −4) with radius 6.
Compare x² = 24y with x² = 4py. Then p = 6, so the receiver belongs 6 units above the vertex, at (0, 6).