Conics

Lessons

Senior 6 · Mathematics

Conics

5 lessons

Lesson 3 of 5

The ellipse

40 min

The curve of the planets

The Earth's orbit round the Sun is an ellipse with the Sun at one focus. NASA's Earth fact sheet gives its closest distance to the Sun (perihelion) as $147.095 \times 10^6\,\mathrm{km}$ and its greatest (aphelion) as $152.100 \times 10^6\,\mathrm{km}$ . An ellipse also explains the "whispering galleries" of some domed buildings, where a whisper at one point is heard clearly across the room at another. Both facts come from the property that defines the curve: the sum of the distances to two fixed points is constant.

Definition

Ellipse: given two points $F_{1}$ and $F_{2}$ , the foci, with $F_{1} F_{2} = 2 c$ , and a length $2a \gt 2c$ , the ellipse is the set of points $P$ with

P F_{1} + P F_{2} = 2 a .

The midpoint of $F_{1} F_{2}$ is the centre. The chord through the foci is the major axis (length $2 a$ ), its ends are the vertices, and the perpendicular chord through the centre is the minor axis (length $2 b$ ). Example: a loop of string of fixed length round two pins traces an ellipse when pulled tight with a pencil.

Important

Standard equation. Take $F_{1} (- c, 0)$ , $F_{2} (c, 0)$ and $P (x, y)$ . Then

\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2} .

Squaring and simplifying gives $a\sqrt{(x - c)^2 + y^2} = a^2 - cx$ . Squaring again gives $(a^{2} - c^{2}) x^{2} + a^{2} y^{2} = a^{2} (a^{2} - c^{2})$ . With $b^{2} = a^{2} - c^{2}$ (so $b \lt a$ ), divide by $a^{2} b^{2}$ :

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 , \qquad c^2 = a^2 - b^2 .

The middle line also says $PF_2 = a - \frac{c}{a}x$ . With the eccentricity $e = \frac{c}{a}$ this is $PF_2 = e\left(\frac{a}{e} - x\right)$ : the distance to $F_{2}$ is $e$ times the distance to the line $x = \frac{a}{e}$ , the directrix of lesson 1.

Element	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , $a \gt b \gt 0$	$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ , $a \gt b \gt 0$
major axis	on the $x$ -axis, length $2 a$	on the $y$ -axis, length $2 a$
vertices	$(\pm a, 0)$	$(0, \pm a)$
ends of minor axis	$(0, \pm b)$	$(\pm b, 0)$
foci	$(\pm c, 0)$ , $c = \sqrt{a^2 - b^2}$	$(0, \pm c)$
eccentricity	$e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}$ , $0 \lt e \lt 1$
directrices	$x = \pm\frac{a}{e}$	$y = \pm\frac{a}{e}$
latus rectum	length $\frac{2b^2}{a}$
parametric form	$(a\cos\theta,\ b\sin\theta)$	$(b\cos\theta,\ a\sin\theta)$

The larger denominator always belongs to the major axis. If $a = b$ , then $c = 0$ , $e = 0$ , and the ellipse is a circle.

The ellipse

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

, drawn from its equation, with both foci, both directrices and the tangent at

P (3, 2.4)

Worked example 1: the elements of an ellipse

Find the vertices, foci, eccentricity, directrices and latus rectum of $16 x^{2} + 9 y^{2} = 144$ .

Solution. Divide by 144: $\frac{x^2}{9} + \frac{y^2}{16} = 1$ . The larger denominator is under $y^{2}$ , so the major axis is vertical, with $a = 4$ and $b = 3$ . Then $c = \sqrt{16 - 9} = \sqrt{7}$ .

vertices $(0, \pm 4)$ ; ends of the minor axis $(\pm 3, 0)$ ; foci $(0, \pm\sqrt{7})$ ;
$e = \frac{\sqrt{7}}{4} \approx 0.661$ ; directrices $y = \pm\frac{a}{e} = \pm\frac{16}{\sqrt{7}} \approx \pm 6.05$ ;
latus rectum $\frac{2b^2}{a} = \frac{18}{4} = 4.5$ .

Worked example 2: the equation from given data

(a) Find the ellipse with foci $(\pm 3, 0)$ and major axis of length 10. (b) Find the ellipse with centre $(1, - 2)$ , a vertex at $(6, - 2)$ and eccentricity $\frac{3}{5}$ .

Solution. (a) $2 a = 10$ , $a = 5$ , $c = 3$ , $b^{2} = 25 - 9 = 16$ : $\frac{x^2}{25} + \frac{y^2}{16} = 1$ .

(b) The vertex is 5 units from the centre along a horizontal line, so $a = 5$ and the major axis is horizontal. $c = a e = 3$ and $b^{2} = 25 - 9 = 16$ : $\frac{(x - 1)^2}{25} + \frac{(y + 2)^2}{16} = 1$ , with foci $(1 \pm 3, -2)$ , that is $(4, - 2)$ and $(- 2, - 2)$ .

Tangent and normal

Differentiating $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ implicitly gives $\frac{dy}{dx} = -\frac{b^2x}{a^2y}$ . At $P (x_{1}, y_{1})$ on the ellipse, the tangent $y - y_1 = -\frac{b^2x_1}{a^2y_1}(x - x_1)$ simplifies (using $\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1$ ) to

\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 , \qquad \text{normal } (x_1y_1 \ne 0):\ \frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 .

Tangent of given slope. The line $y = m x + k$ is a tangent to the ellipse if and only if $k^{2} = a^{2} m^{2} + b^{2}$ . (Substitute and set the discriminant of the quadratic in $x$ to 0.) So for each slope $m$ there are two parallel tangents, $y = mx \pm\sqrt{a^2m^2 + b^2}$ .

Worked example 3: tangent and normal at a point

Show that $P\left(3, \frac{12}{5}\right)$ lies on $\frac{x^2}{25} + \frac{y^2}{9} = 1$ , and find the tangent and normal there.

Solution. $\frac{9}{25} + \frac{144}{225} = \frac{9}{25} + \frac{16}{25} = 1$ . Tangent: $\frac{3x}{25} + \frac{12y}{5 \times 9} = 1$ , that is $\frac{3x}{25} + \frac{4y}{15} = 1$ ; multiplying by 75 gives $9 x + 20 y = 75$ .

Normal: $\frac{25x}{3} - \frac{9y}{\frac{12}{5}} = 25 - 9$ , that is $\frac{25x}{3} - \frac{15y}{4} = 16$ ; multiplying by 12 gives $100 x - 45 y = 192$ . Check: $P$ satisfies it, since $300 - 108 = 192$ , and the slopes $-\frac{9}{20}$ and $\frac{100}{45} = \frac{20}{9}$ multiply to $- 1$ .

The figure also confirms the definition at this point: $PF_1 = \sqrt{7^2 + 2.4^2} = 7.4$ and $PF_2 = \sqrt{1^2 + 2.4^2} = 2.6$ , and $7.4 + 2.6 = 10 = 2 a$ .

Worked example 4: tangents with a given slope

Find the tangents to $\frac{x^2}{25} + \frac{y^2}{9} = 1$ that are parallel to $y = x$ .

Solution. $m = 1$ : $k^{2} = 25 (1) + 9 = 34$ , so the tangents are $y = x + \sqrt{34}$ and $y = x - \sqrt{34}$ .

Worked example 5: the Earth's orbit

NASA gives the Earth's orbit a semi-major axis $a = 149.598 \times 10^6\,\mathrm{km}$ and eccentricity $e = 0.0167$ . The Sun is at a focus. Find the perihelion and aphelion distances and the semi-minor axis, and comment on the shape.

Solution. The Sun is $c = a e$ from the centre. The nearest point of the orbit is the vertex on the Sun's side, at distance $a - c = a(1 - e) \approx 147.10 \times 10^6\,\mathrm{km}$ ; the farthest is $a(1 + e) \approx 152.10 \times 10^6\,\mathrm{km}$ . These agree with NASA's perihelion and aphelion to the precision of $e$ . Then $b = a\sqrt{1 - e^2} \approx 149.58 \times 10^6\,\mathrm{km}$ : $b$ differs from $a$ by about 0.014%, so the orbit is very nearly a circle whose centre is about $2.5 \times 10^6\,\mathrm{km}$ from the Sun.

Important

Reflecting property. The normal at any point $P$ of an ellipse bisects the angle $F_{1} P F_{2}$ . So a ray (of light or sound) leaving one focus is reflected by the ellipse through the other focus. This is the principle of the whispering gallery, and of the medical lithotripter, which focuses shock waves made at one focus of an ellipsoidal reflector onto a kidney stone placed at the other.

Warning

For an ellipse, $c^{2} = a^{2} - b^{2}$ , not $a^{2} + b^{2}$ (that is the hyperbola, lesson 4). A focus is always inside the ellipse, so $c \lt a$ .

Do not assume the major axis is horizontal. In $\frac{x^2}{9} + \frac{y^2}{16} = 1$ the foci are on the $y$ -axis. Compare the denominators first.

Tip

The directrices of an ellipse lie outside it, at $\frac{a}{e} \gt a$ ; the foci lie inside, at $ae \lt a$ . A quick check on any answer: $ae \lt a \lt \frac{a}{e}$ . In the figure, $4 \lt 5 \lt 6.25$ .

Khan Academy: Foci of an ellipse (about 13 min). Watch why the distances from any point to the two foci add up to 2a, and how this gives c² = a² − b².

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Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Find the vertices, foci, eccentricity and directrices of $\frac{x^2}{36} + \frac{y^2}{20} = 1$ .
Find the ellipse with vertices $(\pm 5, 0)$ and eccentricity $\frac{3}{5}$ .
Write $4 x^{2} + y^{2} - 8 x + 4 y - 8 = 0$ in standard form, and give its centre and foci.
Find the tangent and normal to $\frac{x^2}{16} + \frac{y^2}{4} = 1$ at $(2\sqrt{3}, 1)$ .
Find the tangents to $\frac{x^2}{9} + \frac{y^2}{4} = 1$ parallel to $y = 2 x$ .
Mercury's orbit has semi-major axis $57.909 \times 10^6\,\mathrm{km}$ and eccentricity 0.2056 (NASA). Find its perihelion and aphelion distances to 3 significant figures.
Using the directrices, prove that $P F_{1} + P F_{2} = 2 a$ for every point $P$ of $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ .

Show answersHide answers

$a = 6$ , $b = 2\sqrt{5}$ , $c = 4$ : vertices $(\pm 6, 0)$ , foci $(\pm 4, 0)$ , $e = \frac{2}{3}$ , directrices $x = \pm 9$ .
$c = 3$ , $b = 4$ : $\frac{x^2}{25} + \frac{y^2}{16} = 1$ .
$\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{16} = 1$ : centre $(1, - 2)$ , $c = 2\sqrt{3}$ , foci $(1, -2 \pm 2\sqrt{3})$ .
Tangent $\sqrt{3}x + 2y = 8$ ; normal $2x - \sqrt{3}y = 3\sqrt{3}$ .
$k^{2} = 9 (4) + 4 = 40$ : $y = 2x \pm 2\sqrt{10}$ .
Perihelion $46.0 \times 10^6\,\mathrm{km}$ , aphelion $69.8 \times 10^6\,\mathrm{km}$ .
$PF_2 = e\left(\frac{a}{e} - x\right) = a - ex$ and $PF_1 = e\left(x + \frac{a}{e}\right) = a + ex$ , so the sum is $2 a$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the diagram (plotted from $\frac{x^2}{25} + \frac{y^2}{9} = 1$ with the computed tangent), the worked examples, the practice items and the quiz questions.
Orbit data: Earth Fact Sheet and Mercury Fact Sheet, NASA Space Science Data Coordinated Archive (US government work, public domain).
Video: Foci of an ellipse, Khan Academy, embedded from YouTube.

1The curve of the planets

2Tangent and normal

The curve of the planets

Tangent and normal