Formulas · Mathematics

Conic sections

7 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 11

Radius of a circle from its centre and a point

r = (x - h)^{2} + (y - k)^{2}

What each letter means
$r$	the radius of the circle
$h$	the x-coordinate of the centre
$k$	the y-coordinate of the centre
$x$	the x-coordinate of a point on the circle
$y$	the y-coordinate of a point on the circle

The circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r²: every point on it is r away from the centre.

Why it works

A point (x, y) is on the circle exactly when its distance from the centre is r. By the distance formula that distance is √((x − h)² + (y − k)²); squaring both sides gives the equation.

When to use it

Writing the equation of a circle from its centre and radius, finding the radius from the centre and one point, reading the centre and radius from an equation, and testing whether a point is inside, on or outside a circle.

How to use it

Put the centre into (x − h)² + (y − k)² = r², watching the signs.
To read an equation, the centre's coordinates have the opposite signs of the numbers in the brackets.

To remember: A circle is every point at one distance from the centre.

Other forms

Same formula, another form $(x - h)^{2} + (y - k)^{2} = r^{2}$ The equation of the circle, as the book writes it.
Special case $x^{2} + y^{2} = r^{2}$ When the centre is the origin (h = k = 0).

Worked examples

Story

A garden sprinkler at (1, 1) just reaches a flower bed at (7, 9), in metres. How far does it spray?

$(7 - 1)^{2} + (9 - 1)^{2} = 100$
$100 = 10$

Answer: 10

Picture

Find the equation of the circle with centre (−3, 2) and radius 4.

(x + 3)² + (y − 2)² = 16. Multiplied out, that is x² + y² + 6x − 4y − 3 = 0.
$9 + 4 - 16 = - 3$

Answer: (x + 3)² + (y − 2)² = 16

Direct

A circle has centre (2, 3) and passes through (5, 7). Find its radius.

$(5 - 2)^{2} + (7 - 3)^{2} = 25$
$25 = 5$

Answer: 5

Reverse

A circle with centre (0, 0) and radius 13 passes through (5, y), with y positive. Find y.

Try it first, then show the working

$5^{2} + 1 2^{2} = 13$

Answer: 12

Exam

Does the point (2, 3) lie inside, on or outside the circle x² + y² = 16?

Try it first, then show the working

Find its distance from the centre (0, 0):
$2^{2} + 3^{2} = 13$
The radius is 4 and √13 is less than 4, so the point is inside.

Answer: Inside

Common mistake: Reading (x + 2)² as a centre at x = 2 (it is x = −2), and forgetting to square the radius.

Class 11

Radius from the general equation of a circle

r = g^{2} + f^{2} - c

What each letter means
$r$	the radius of the circle
$g$	half the number beside x in x² + y² + 2gx + 2fy + c = 0
$f$	half the number beside y
$c$	the constant term

The equation x² + y² + 2gx + 2fy + c = 0 is a circle with centre (−g, −f) and radius √(g² + f² − c).

Why it works

Complete the squares: x² + 2gx = (x + g)² − g² and y² + 2fy = (y + f)² − f². So the equation becomes (x + g)² + (y + f)² = g² + f² − c, the centre-radius form with centre (−g, −f).

When to use it

Finding the centre and radius when a circle's equation has been multiplied out.

How to use it

Make the numbers in front of x² and y² equal to 1 (divide through).
Halve the numbers beside x and y to get g and f.
The centre is (−g, −f) and the radius is √(g² + f² − c).

To remember: Halve and flip the signs for the centre; square, add and take away c for r².

Other forms

Special case $r = g^{2} + f^{2}$ When the circle passes through the origin (c = 0).
To remember itIf g² + f² − c is 0, the "circle" is a single point; if it is negative, there is no circle at all.

Worked examples

Story

On a map, the edge of a round pond is x² + y² + 8x − 2y − 8 = 0, in metres. What is its radius?

$4^{2} + (- 1)^{2} + 8 = 5$

Answer: 5

Picture

Find the centre and radius of 2x² + 2y² − 8x − 10 = 0.

Divide by 2: x² + y² − 4x − 5 = 0, so g = −2, f = 0 and c = −5.
$4 + 0 + 5 = 3$
The centre is (2, 0) and the radius is 3.

Answer: Centre (2, 0), radius 3

Direct

Find the radius of x² + y² − 4x + 6y − 12 = 0.

Here 2g = −4 and 2f = 6, so g = −2, f = 3, and c = −12.
$(- 2)^{2} + 3^{2} + 12 = 25$
$25 = 5$

Answer: 5

Reverse

The circle x² + y² − 6x + 4y + c = 0 has radius 4. Find c.

Try it first, then show the working

$9 + 4 - (- 3) = 16$

Answer: -3

Exam

Find the equation of the circle through the origin with centre (3, 4).

Try it first, then show the working

Through the origin means c = 0; the centre (3, 4) means g = −3 and f = −4.
$9 + 16 = 5$
The circle is x² + y² − 6x − 8y = 0.

Answer: x² + y² − 6x − 8y = 0

Common mistake: Forgetting to halve the numbers beside x and y, or keeping their signs for the centre.

Class 11

Latus rectum of a parabola

L = 4 a

What each letter means
$L$	the length of the latus rectum: the chord through the focus, at right angles to the axis
$a$	the distance from the vertex to the focus, in y² = 4ax

The parabola y² = 4ax has its vertex at the origin, its focus at (a, 0), its directrix x = −a, and a latus rectum 4a long.

Why it works

The latus rectum passes through the focus, where x = a. Putting x = a into y² = 4ax gives y² = 4a², so y = 2a or −2a: the chord runs from (a, −2a) to (a, 2a), which is 4a long.

When to use it

Reading the focus, directrix and latus rectum from a parabola's equation, and writing the equation from them.

How to use it

Match the equation to y² = 4ax (or x² = 4ay): the number beside x is 4a.
Divide it by 4 to get a.
Focus (a, 0), directrix x = −a, latus rectum 4a.

To remember: Four a across the focus.

Other forms

To remember itFor x² = 4ay, x and y swap roles: focus (0, a), directrix y = −a, and the latus rectum is still 4a.

Worked examples

Story

A satellite dish is shaped like y² = 20x (in cm) through its middle, with the receiver at the focus. How long is the chord of the dish through the receiver, straight across the axis?

$4 \times 5 = 20$

Answer: 20

Picture

For y² = 8x, find the focus, the directrix and the ends of the latus rectum.

Here 4a = 8, and so a = 2: the focus is (2, 0) and the directrix is x = −2.
At x = 2, y² = 16:
$4 \times 2 \times 2 = 4$

Answer: Focus (2, 0), directrix x = −2, ends (2, 4) and (2, −4)

Direct

Find the length of the latus rectum of y² = 12x.

Here 4a = 12, and so a = 3.
$4 \times 3 = 12$

Answer: 12

Reverse

A parabola y² = 4ax has a latus rectum 6 long. Find a and the equation.

Try it first, then show the working

$4 \times 1.5 = 6$
The parabola is y² = 6x.

Answer: 1.5

Exam

Find the equation of the parabola with its vertex at the origin and its focus at (0, −3).

Try it first, then show the working

The focus is on the negative y-axis, so the form is x² = −4ay, with a = 3.
$4 \times 3 = 12$
The parabola is x² = −12y.

Answer: x² = −12y

Common mistake: Taking a as the whole number beside x: in y² = 12x, 4a = 12, so a = 3, not 12.

Class 11

Foci of an ellipse

c = a^{2} - b^{2}

What each letter means
$c$	the distance from the centre to each focus
$a$	the semi-major axis: half the longer axis
$b$	the semi-minor axis: half the shorter axis

For the ellipse x²/a² + y²/b² = 1, with a larger than b, the foci are at (c, 0) and (−c, 0), where c² = a² − b².

Why it works

The end of the minor axis, (0, b), is on the ellipse, so its distances to the two foci add up to 2a; by symmetry each is a. That point, the centre and a focus make a right triangle with hypotenuse a and legs b and c, so a² = b² + c².

When to use it

Finding the foci from the equation, and finding a or b from the foci and one axis.

How to use it

Read a² and b² under x² and y²; a is the larger.
Work out c = √(a² − b²).
The foci lie on the major axis, c from the centre.

To remember: Ellipse: the foci are inside, so c is the smaller: c² = a² − b².

Other forms

Rearranged $c^{2} = a^{2} - b^{2}$ The same fact without the square root.
Rearranged $b = a^{2} - c^{2}$ The semi-minor axis from a and c.

Worked examples

Story

An elliptical running track is 100 m long and 60 m wide. How far from the centre are its foci?

$5 0^{2} - 3 0^{2} = 1600$
$1600 = 40$

Answer: 40

Picture

Find the foci and the lengths of the axes of x²/16 + y²/25 = 1.

25 is under y², so the major axis is along the y-axis: a = 5 and b = 4.
$25 - 16 = 3$
The foci are (0, 3) and (0, −3); the axes are 10 and 8 long.

Answer: Foci (0, ±3); axes 10 and 8

Direct

For x²/25 + y²/9 = 1, find c, the distance from the centre to each focus, and name the foci.

$25 - 9 = 4$
The foci are (4, 0) and (−4, 0).

Answer: 4

Reverse

An ellipse has foci (±3, 0) and a major axis 10 long. Find b.

Try it first, then show the working

$25 - 16 = 3$
So b² = 16 and b = 4: the ellipse is x²/25 + y²/16 = 1.

Answer: 4

Exam

Find the equation of the ellipse with vertices (±13, 0) and foci (±5, 0).

Try it first, then show the working

Here a = 13 and c = 5, and b² = a² − c².
$169 - 25 = 144$
The ellipse is x²/169 + y²/144 = 1.

Answer: x²/169 + y²/144 = 1

Common mistake: Adding instead of taking away (that is the hyperbola), or taking a as the number under x² when the larger one is under y² (then the major axis is vertical).

Class 11

Foci of a hyperbola

c = a^{2} + b^{2}

What each letter means
$c$	the distance from the centre to each focus
$a$	the semi-transverse axis: from the centre to a vertex
$b$	the semi-conjugate axis

For the hyperbola x²/a² − y²/b² = 1 the foci are at (c, 0) and (−c, 0), where c² = a² + b².

Why it works

On a hyperbola the foci lie beyond the vertices, so c is more than a. The book names b by b² = c² − a², which turns the defining property (the difference of the distances to the foci is 2a) into x²/a² − y²/b² = 1.

When to use it

Finding the foci of a hyperbola from its equation, or its equation from the foci and vertices.

How to use it

Read a² under the positive term and b² under the negative one.
Work out c = √(a² + b²).
The foci lie on the axis of the positive term.

To remember: Hyperbola: the foci are outside, so c is the bigger: c² = a² + b².

Other forms

Rearranged $c^{2} = a^{2} + b^{2}$ The same fact without the square root.
Rearranged $b = c^{2} - a^{2}$ The semi-conjugate axis from a and c.

Worked examples

Story

The side of a cooling tower follows x²/36 − y²/64 = 1, in metres. How far from the centre are the foci of this curve?

$36 + 64 = 10$

Answer: 10

Picture

Find the foci and the vertices of y²/9 − x²/27 = 1.

The y² term is positive, so the axis is the y-axis: a = 3 and b² = 27.
$9 + 27 = 6$
The foci are (0, 6) and (0, −6); the vertices are (0, 3) and (0, −3).

Answer: Foci (0, ±6); vertices (0, ±3)

Direct

For x²/16 − y²/9 = 1, find c and the foci.

$16 + 9 = 5$
The foci are (5, 0) and (−5, 0).

Answer: 5

Reverse

A hyperbola has foci (±5, 0) and vertices (±3, 0). Find b.

Try it first, then show the working

$9 + 16 = 5$

Answer: 4

Exam

Find the equation of the hyperbola with foci (0, ±13) and vertices (0, ±5).

Try it first, then show the working

The foci are on the y-axis, with a = 5 and c = 13; then b² = c² − a².
$169 - 25 = 144$
The hyperbola is y²/25 − x²/144 = 1.

Answer: y²/25 − x²/144 = 1

Common mistake: Taking away as for an ellipse, or taking the larger number as a²: for a hyperbola, a² is under the positive term, whichever is larger.

Class 11

Eccentricity of an ellipse or hyperbola

e = \frac{c}{a}

What each letter means
$e$	the eccentricity: how stretched the curve is
$c$	the distance from the centre to a focus
$a$	the distance from the centre to a vertex (the semi-major or semi-transverse axis)

The eccentricity e = c ÷ a measures how stretched a conic is: 0 for a circle, between 0 and 1 for an ellipse, more than 1 for a hyperbola.

Why it works

In an ellipse c is less than a, so e is less than 1; as the foci move out towards the vertices, e grows towards 1 and the ellipse flattens. In a hyperbola c is more than a, so e is more than 1. A circle has both foci at the centre, so c = 0 and e = 0.

When to use it

Describing the shape of an ellipse or hyperbola, and finding c (and so the foci) from e and a.

How to use it

Find c from a and b: c² = a² − b² for an ellipse, a² + b² for a hyperbola.
Divide c by a.

To remember: Circle 0, ellipse under 1, hyperbola over 1.

Other forms

Rearranged $c = a e$ The focus distance from e and a.
Special case $e = 1 - \frac{b ^{2}}{a ^{2}}$ When the curve is an ellipse (c² = a² − b²).

Worked examples

Story

The Earth's orbit is an ellipse with a about 150 million km and c about 2.5 million km. Find its eccentricity.

$\frac{2.5}{150} = \frac{1}{60}$
That is about 0.017: the orbit is very nearly a circle.

Answer: 0.016666666666666666

Picture

Find the eccentricity of the hyperbola x²/9 − y²/16 = 1.

First c = √(9 + 16):
$9 + 16 = 5$
Then e = 5/3, more than 1, as for every hyperbola.

Answer: 1.6666666666666667

Direct

Find the eccentricity of x²/25 + y²/9 = 1, where c = 4.

$\frac{4}{5} = 0.8$

Answer: 0.8

Reverse

An ellipse has eccentricity ½ and a = 8. How far from the centre are its foci?

Try it first, then show the working

$8 \times 0.5 = 4$

Answer: 4

Exam

Find the equation of the ellipse with foci (±2, 0) and eccentricity ½.

Try it first, then show the working

Here a = c ÷ e = 2 ÷ ½ = 4, and b² = a² − c².
$16 - 4 = 12$
The ellipse is x²/16 + y²/12 = 1.

Answer: x²/16 + y²/12 = 1

Common mistake: Dividing by b, or using the ellipse rule for c on a hyperbola.

Class 11

Latus rectum of an ellipse or hyperbola

L = \frac{2 b ^{2}}{a}

What each letter means
$L$	the length of the latus rectum: the chord through a focus, at right angles to the axis the foci lie on
$a$	the semi-axis along which the foci lie
$b$	the other semi-axis

The latus rectum of the ellipse x²/a² + y²/b² = 1, and of the hyperbola x²/a² − y²/b² = 1, is 2b² ÷ a long.

Why it works

The latus rectum is the chord through the focus (c, 0) at right angles to the axis. For the ellipse, at x = c, y²/b² = 1 − c²/a² = b²/a², so y = b²/a or −b²/a and the chord is 2b²/a. The hyperbola works the same way with c² = a² + b².

When to use it

Sketching an ellipse or hyperbola (the latus rectum gives four more points), and exam questions that give its length.

How to use it

Read a (the semi-axis the foci are on) and b.
Work out 2b² ÷ a.

To remember: Two b squared over a.

Other forms

To remember itFor a parabola y² = 4ax the latus rectum is 4a instead.

Worked examples

Story

An elliptical mirror follows x²/36 + y²/9 = 1, in cm, with a lamp at a focus. How long is the chord of the mirror through the lamp, at right angles to the long axis?

$\frac{2 \times 9}{6} = 3$

Answer: 3

Picture

Find the ends of the latus rectum through the right-hand focus of x²/9 − y²/16 = 1.

First c = √(9 + 16) = 5, so that focus is (5, 0).
Half the latus rectum is b² ÷ a = 16/3.
$\frac{2 \times 16}{3} = \frac{32}{3}$

Answer: (5, 16/3) and (5, −16/3)

Direct

Find the length of the latus rectum of x²/25 + y²/16 = 1.

$\frac{2 \times 16}{5} = 6.4$

Answer: 6.4

Reverse

An ellipse with a = 10 has a latus rectum 5 long. Find b.

Try it first, then show the working

$\frac{2 \times 25}{10} = 5$

Answer: 5

Exam

The latus rectum of an ellipse is half its major axis. Find its eccentricity.

Try it first, then show the working

The rule 2b² ÷ a = a gives b² = a² ÷ 2.
Then e² = 1 − b²/a² = 1 − ½.
$1 - \frac{1}{2} = \frac{1}{2}$

Answer: 1/√2

Common mistake: Writing 2a² ÷ b, or taking a from the wrong axis.