Conic Sections

A cone here means a double-napped right circular cone: two identical nappes joined at a common vertex, extending infinitely along a fixed line called the axis. Every line on the cone's surface through the vertex is a generator, making a fixed angle α (the semi-vertical angle) with the axis.

When a plane cuts the cone (not through the vertex), the shape of the intersection depends on the angle β between the cutting plane and the axis, compared with α:

• If β = 90° (plane perpendicular to the axis), the section is a circle.
• If α < β < 90°, the section is an ellipse.
• If β = α (plane parallel to a generator), the section is a parabola.
• If 0 ≤ β < α, the plane cuts both nappes and the section is a hyperbola (two branches).

If the cutting plane passes through the vertex, the section degenerates into a single point, a single straight line, or a pair of intersecting straight lines. These are the degenerate conics.

Worked example: A plane perpendicular to the axis of a cone cuts it above the vertex (not through it), giving a circle since β = 90°. If the same plane is tilted so that α < β < 90° still holds, the circle stretches into an ellipse. This shows the circle is really the special (limiting) case of an ellipse in which both foci coincide at the centre.

Quick answer: A circle is the set of all points in a plane at a constant distance (the radius, r) from a fixed point (the centre).

If the centre is C(h, k) and P(x, y) is any point on the circle, then CP = r. Squaring the distance formula gives the standard equation of a circle: (x - h)² + (y - k)² = r². When the centre is the origin, this reduces to x² + y² = r².

Expanding the standard equation gives the general form x² + y² + 2gx + 2fy + c = 0. Comparing coefficients, the centre of this general equation is (-g, -f) and the radius is √(g² + f² - c) (real and positive only when g² + f² > c).

Worked example 1: Find the equation of the circle with centre (-3, 2) and radius 4. Using (x - h)² + (y - k)² = r² with h = -3, k = 2, r = 4, the equation is (x + 3)² + (y - 2)² = 16.

Worked example 2: Find the centre and radius of the circle x² + y² - 4x + 6y - 12 = 0. Comparing with x² + y² + 2gx + 2fy + c = 0 gives 2g = -4 so g = -2, 2f = 6 so f = 3, and c = -12. Hence centre = (-g, -f) = (2, -3), and radius = √(g² + f² - c) = √(4 + 9 + 12) = √25 = 5.

Quick answer: A parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix) not containing the focus.

The line through the focus perpendicular to the directrix is the axis; the point where the parabola meets its axis is the vertex (midway between focus and directrix). The chord through the focus, perpendicular to the axis, is the latus rectum, measuring how wide the parabola is.

Taking the vertex at the origin, axis along the x-axis, focus at (a, 0), a > 0, and directrix x = -a, the equidistance condition simplifies to the standard equation y² = 4ax. There are four standard forms, each with vertex at the origin:

• y² = 4ax opens right; focus (a, 0), directrix x = -a.
• y² = -4ax opens left; focus (-a, 0), directrix x = a.
• x² = 4ay opens upward; focus (0, a), directrix y = -a.
• x² = -4ay opens downward; focus (0, -a), directrix y = a.

In every case the length of the latus rectum is 4a, and for y² = 4ax its endpoints are (a, 2a) and (a, -2a).

Worked example: Find the focus, axis, directrix and length of the latus rectum of y² = 12x. Comparing with y² = 4ax gives 4a = 12, so a = 3. Hence focus = (3, 0), axis is y = 0, directrix is x = -3, and the length of the latus rectum is 4a = 12.

Quick answer: An ellipse is the set of all points the sum of whose distances from two fixed points (the foci) is a constant, 2a.

Let the foci be F₁(-c, 0) and F₂(c, 0), with constant sum 2a (a > c > 0). Writing b² = a² - c² and simplifying PF₁ + PF₂ = 2a gives the standard equation with centre at the origin, major axis along the x-axis: x²/a² + y²/b² = 1, with a > b > 0.

Here the vertices are (±a, 0), the foci are (±c, 0) where c² = a² - b², the major axis has length 2a, and the minor axis has length 2b. The eccentricity e = c/a (0 < e < 1) measures how stretched the ellipse is; also b² = a²(1 - e²).

The focal chord perpendicular to the major axis is the latus rectum, of length 2b²/a. If instead the larger denominator is under y², the major axis lies along the y-axis, and the foci become (0, ±c).

Worked example: Find the foci, vertices, lengths of the major and minor axes, and eccentricity of x²/16 + y²/9 = 1. Here a² = 16, b² = 9 (major axis along x-axis), so a = 4, b = 3. Then c² = a² - b² = 16 - 9 = 7, so c = √7. Hence foci = (±√7, 0), vertices = (±4, 0), major axis length = 8, minor axis length = 6, and e = c/a = √7/4.

Quick answer: A hyperbola is the set of all points the absolute difference of whose distances from two fixed foci is a constant, 2a.

Let the foci be F₁(-c, 0) and F₂(c, 0), with |PF₁ - PF₂| = 2a (c > a > 0). Writing b² = c² - a² gives the standard equation centred at the origin, transverse axis along the x-axis: x²/a² - y²/b² = 1.

Here the vertices are (±a, 0), the foci are (±c, 0) with c² = a² + b², the transverse axis has length 2a, and the conjugate axis has length 2b. The eccentricity e = c/a is always greater than 1, and b² = a²(e² - 1).

The latus rectum is the focal chord perpendicular to the transverse axis, of length 2b²/a. If the transverse axis lies along the y-axis instead, the standard form becomes y²/a² - x²/b² = 1, with foci (0, ±c).

Worked example: Find the eccentricity, foci and length of the latus rectum of x²/9 - y²/16 = 1. Here a² = 9, b² = 16, so a = 3, b = 4. Then c² = a² + b² = 9 + 16 = 25, so c = 5. Hence e = c/a = 5/3, foci = (±5, 0), and latus rectum length = 2b²/a = 2(16)/3 = 32/3.

Quick answer: Every conic section can be classified purely by its eccentricity e, and every standard formula shifts predictably once the centre or vertex is moved from the origin to a general point (h, k).

The eccentricity e is the single number that separates the four conics:

• e = 0: circle (both foci coincide at the centre).
• 0 < e < 1: ellipse.
• e = 1: parabola.
• e > 1: hyperbola.

When the axes of the conic remain parallel to the coordinate axes but the vertex or centre is shifted to (h, k), every standard equation simply has x replaced by (x - h) and y by (y - k):

• Circle: (x - h)² + (y - k)² = r².
• Parabola (opening right, vertex (h, k)): (y - k)² = 4a(x - h).
• Ellipse (major axis horizontal, centre (h, k)): (x - h)²/a² + (y - k)²/b² = 1.
• Hyperbola (transverse axis horizontal, centre (h, k)): (x - h)²/a² - (y - k)²/b² = 1.

Worked example: Identify the conic 9x² + 4y² = 36 and find its eccentricity. Dividing by 36 gives x²/4 + y²/9 = 1. Since the denominator under y² (9) is larger, this is an ellipse with major axis along the y-axis, a² = 9, b² = 4, so a = 3, b = 2. Then c² = a² - b² = 9 - 4 = 5, c = √5, and e = c/a = √5/3, confirming 0 < e < 1 as expected for an ellipse.