Generalities on conic sections – Conics (Senior 6 Mathematics) | Ganzaa

Conics

Lessons

Senior 6 · Mathematics

Conics

5 lessons

Lesson 1 of 5

Generalities on conic sections

35 min

Four curves from one cone

About 200 BC the Greek geometer Apollonius of Perga wrote eight books on the curves you get by slicing a cone, and gave them the names we still use: ellipse, parabola and hyperbola. For eighteen centuries they were pure geometry. Then in 1609 Johannes Kepler showed that Mars moves round the Sun in an ellipse, and later Isaac Newton proved that every body moving under gravity follows a conic. Today a satellite dish, a car headlight and the radio navigation once used by ships are all built on the same three curves.

In this unit you meet each conic in three ways: as a slice of a cone, as a curve defined by distances to a point and a line, and as an equation of the second degree in $x$ and $y$ .

Definition

Double cone: the surface traced by a line (a generator) turning about a fixed axis that it crosses at a fixed point, the vertex, at a constant angle. The two halves on either side of the vertex are the nappes.

Conic section (or conic): the curve in which a plane not through the vertex cuts a double cone. If the plane

cuts every generator of one nappe, the conic is an ellipse (a circle when the plane is perpendicular to the axis);
is parallel to exactly one generator, it is a parabola;
cuts both nappes, it is a hyperbola, which has two branches.

A plane through the vertex gives a degenerate conic: a single point, a line, or a pair of crossing lines.

A double cone seen from the side. Each cutting plane is seen edge-on as a line; the colour of the line names the curve it cuts out.

Definition

Focus–directrix definition. Fix a line $d$ , the directrix, a point $F$ not on it, the focus, and a number $e \gt 0$ , the eccentricity. The conic is the set of points $P$ such that

PF = e \cdot PM ,

where $P M$ is the distance from $P$ to the directrix ( $M$ is the foot of the perpendicular). The conic is an ellipse if $0 \lt e \lt 1$ , a parabola if $e = 1$ , and a hyperbola if $e \gt 1$ . A circle is the limiting case $e = 0$ : it has a centre but no directrix.

The line through $F$ perpendicular to the directrix is the focal axis; the conic is symmetric about it.

Same focus, same directrix, three eccentricities. Each curve is drawn from its equation

r = \frac{2e}{1 - e\cos\theta}

(lesson 5); only the branch of the hyperbola nearer F appears.

Worked example 1: a conic from its focus and directrix

Find the equation of the conic with focus $F (1, 0)$ , directrix $x = 4$ and eccentricity $e = \frac{1}{2}$ , and name it.

Solution. For $P (x, y)$ : $PF = \sqrt{(x - 1)^2 + y^2}$ and $P M = ∣ x - 4 ∣$ . Square $PF = \frac{1}{2}PM$ :

(x - 1)^2 + y^2 = \frac{1}{4}(x - 4)^2 \iff 4x^2 - 8x + 4 + 4y^2 = x^2 - 8x + 16 \iff 3x^2 + 4y^2 = 12 .

Dividing by 12: $\frac{x^2}{4} + \frac{y^2}{3} = 1$ . Since $e = \frac{1}{2} \lt 1$ , this is an ellipse, centred at the origin, with $a = 2$ along the $x$ -axis.

The general equation of the second degree

Expanding $PF^2 = e^2 \, PM^2$ always gives an equation of the form

Ax^2 + Bxy + Cy^2 + Dx + Ey + G = 0 , \qquad A, B, C \text{ not all } 0 .

(Some books call the constant term $F$ ; here $F$ is kept for the focus.) Conversely, every such equation is a conic, a degenerate conic, or has no points at all. A rotation of the axes removes the $x y$ term without changing the type of curve, and the type can be read from the discriminant $B^{2} - 4 A C$ :

$B^{2} - 4 A C$	Type (if not degenerate)	When $B = 0$	Example
$\lt 0$	ellipse (circle if $B = 0$ and $A = C$ )	$A$ and $C$ have the same sign	$4 x^{2} + 9 y^{2} = 36$
$= 0$	parabola	$A = 0$ or $C = 0$	$y^{2} = 8 x$
$\gt 0$	hyperbola	$A$ and $C$ have opposite signs	$x^{2} - 4 y^{2} = 4$

When $B = 0$ , complete the square in $x$ and in $y$ to find the centre (or vertex) and the standard form.

Worked example 2: identify by completing the square

Name each conic and give its centre or vertex. (a) $4 x^{2} + 9 y^{2} - 16 x + 18 y - 11 = 0$ (b) $y^{2} - 6 y - 8 x + 25 = 0$ (c) $x^{2} - 4 y^{2} + 2 x + 16 y - 19 = 0$

Solution. (a) $AC = 36 \gt 0$ : an ellipse. $4 (x^{2} - 4 x) + 9 (y^{2} + 2 y) = 11$ , so $4 (x - 2)^{2} - 16 + 9 (y + 1)^{2} - 9 = 11$ , that is $4 (x - 2)^{2} + 9 (y + 1)^{2} = 36$ :

\frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} = 1 , \quad \text{centre } (2, -1),\ a = 3,\ b = 2 .

(b) $A = 0$ : a parabola. $(y - 3)^{2} - 9 = 8 x - 25$ , so $(y - 3)^{2} = 8 x - 16 = 8 (x - 2)$ : vertex $(2, 3)$ , opening to the right.

(c) $AC = -4 \lt 0$ : a hyperbola. $(x + 1)^2 - 1 - 4\big[(y - 2)^2 - 4\big] - 19 = 0$ , so $(x + 1)^{2} - 4 (y - 2)^{2} = 4$ :

\frac{(x + 1)^2}{4} - (y - 2)^2 = 1 , \quad \text{centre } (-1, 2) .

Worked example 3: the discriminant with an $x y$ term

Classify (a) $3 x^{2} + 2 x y + 3 y^{2} - 8 = 0$ (b) $x^{2} + 4 x y + 4 y^{2} - 6 x = 0$ (c) $x y = 4$ .

Solution. (a) $B^2 - 4AC = 4 - 36 = -32 \lt 0$ : an ellipse. (b) $16 - 16 = 0$ : a parabola. (c) Written $x y - 4 = 0$ : $A = C = 0$ , $B = 1$ , so $B^2 - 4AC = 1 \gt 0$ : a hyperbola, with the axes as asymptotes (lesson 4).

Warning

The discriminant does not detect degenerate cases. $x^{2} - 9 y^{2} = 0$ has $B^2 - 4AC = 36 \gt 0$ but factorises as $(x - 3 y) (x + 3 y) = 0$ : two crossing lines. $x^{2} + 2 y^{2} = 0$ is only the point $(0, 0)$ , and $x^{2} + y^{2} + 1 = 0$ has no real points. After completing the square, check that the right-hand side has the sign the standard form needs.

Tip

To remember which eccentricity is which: an ellipse "falls short" ( $e \lt 1$ ), a parabola is "equal" ( $e = 1$ ) and a hyperbola "overshoots" ( $e \gt 1$ ). The Greek words elleipsis, parabole and hyperbole mean exactly that.

3Blue1Brown: Why slicing a cone gives an ellipse (about 13 min). Watch how two spheres touching the cone and the cutting plane (Dandelin spheres) prove that the slice has two foci whose distances add up to a constant.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks you to explain.

Use $B^{2} - 4 A C$ to classify (a) $x^{2} + x y + y^{2} - 3 = 0$ (b) $x^{2} + 2 x y + y^{2} + x - y = 0$ (c) $2 x^{2} + 5 x y + 2 y^{2} = 1$ .
Write $9 x^{2} + 4 y^{2} + 36 x - 8 y + 4 = 0$ in standard form. Name the conic and give its centre.
Write $x^{2} + 4 x - 4 y + 16 = 0$ in standard form, and give the vertex and the direction in which the curve opens.
Find the equation of the conic with focus $(2, 0)$ , directrix $x = - 1$ and eccentricity 2, and name it.
Find the conic with focus $(0, 1)$ and directrix $y = - 1$ and eccentricity 1.
For which real $k$ is $x^{2} + k y^{2} = 4$ (a) a circle (b) an ellipse that is not a circle (c) a hyperbola (d) degenerate?
Explain why a plane through the vertex of a double cone cannot cut out an ellipse with two distinct points on it.

Show answersHide answers

(a) $1 - 4 = - 3$ : ellipse (b) $4 - 4 = 0$ : parabola (c) $25 - 16 = 9$ : hyperbola
$\frac{(x + 2)^2}{4} + \frac{(y - 1)^2}{9} = 1$ : an ellipse with centre $(- 2, 1)$ , major axis vertical.
$(x + 2)^{2} = 4 (y - 3)$ : vertex $(- 2, 3)$ , opening upwards.
$(x - 2)^{2} + y^{2} = 4 (x + 1)^{2}$ gives $3 (x + 2)^{2} - y^{2} = 12$ , that is $\frac{(x + 2)^2}{4} - \frac{y^2}{12} = 1$ : a hyperbola.
$x^{2} + (y - 1)^{2} = (y + 1)^{2}$ gives $x^{2} = 4 y$ : a parabola.
(a) $k = 1$ (b) $k \gt 0$ , $k \ne 1$ (c) $k \lt 0$ (d) $k = 0$ : the pair of lines $x = \pm 2$ .
Every point of the cone on the plane lies on a generator through the vertex, and the plane meets the cone in whole generators (or only in the vertex). So the section is a point, one line or two lines, never a closed curve.

Sources and credits

Written by Ganzaa: the explanations, both diagrams (the conics plotted from their polar equations), the worked examples, the practice items and the quiz questions. Every standard form was checked by expanding it back.
Video: Why slicing a cone gives an ellipse (beautiful proof), 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1Four curves from one cone

2The general equation of the second degree

Four curves from one cone

The general equation of the second degree