Download Conics

Academic Skills Advice
Conics
The main shapes known as conics are:
Parabola (also known as quadratics as looked at in the last lesson)
Ellipse (including the circle which is a special case)
Hyperbola (including the rectangular hyperbola which is a special case)
Some of them include a 𝑦 2 term as well as (or instead of) an 𝑥 2 term
You can recognise a conic from its general equation. The general equation tells you enough
about the shape to enable you to sketch it.
The general equations are as follows:
(𝒙 − 𝒂)𝟐 + (𝒚 − 𝒃)𝟐 = 𝒓𝟐
Circle:
Ellipse:
Hyperbola:
𝑥2
𝑦2
𝑎
𝑏2
+
2
𝑥2
𝑦2
𝑎
𝑏2
−
2
=1
centre = (𝒂,𝒃) and radius= 𝒓
𝑥-intercepts = ±𝑎
𝑦-intercepts = ±𝑏
=1
Points to note:
Circle:
If 𝑎 and 𝑏 are both 0, the equation becomes 𝑥 2 + 𝑦 2 = 𝑟 2 (i.e. a circle with
centre at (0, 0) and radius r.)
To find the centre and the radius, the coefficients of 𝑥 2 and 𝑦 2 must = 1.
Ellipse:
The equation must = 1. (i.e. you need 1 on the right hand side).
You need squared numbers on the bottom of each fraction.
The squared number beneath 𝑥 tells us the 𝑥-intercepts and the squared
number beneath 𝑦 tells us the 𝑦-intercepts.
Hyperbola: The equation must = 1. (i.e. you need 1 on the right hand side).
You need squared numbers on the bottom of each fraction.
The squared number beneath 𝑥 tells us where to draw the vertical edges of the
rectangle and the squared number beneath 𝑦 tells us where to draw the
horizontal edges of the rectangle.
© H Jackson 2011 / ACADEMIC SKILLS
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Circle
Notice that for a circle:
 the 𝑥 and 𝑦 terms are both squared, then added.
 the coefficients of 𝑥 2 and 𝑦 2 are equal
 there is no 𝑥𝑦 term.
(𝒙 − 𝒂)𝟐 + (𝒚 − 𝒃)𝟐 = 𝒓𝟐
centre = (𝒂,𝒃) and radius= 𝒓
Examples:

Sketch the circle with equation (𝒙 − 𝟑)𝟐 + (𝒚 + 𝟐)𝟐 = 𝟏𝟔
From the equation we see that the centre of the circle is at (3, −2) and the radius is 4
(because √16 = 4)
𝑦
Notice that the co-ordinates of the
centre have the opposite signs to
the numbers in the brackets
2
𝑥
-1
x
(3, −2)

Sketch the circle with equation 𝟑𝒙𝟐 + 𝟑𝒚𝟐 = 𝟏𝟐
We need the coefficients of 𝑥 2 and 𝑦 2 to equal 1 so we divide the equation by 3.
Starting equation:
÷3:
We now have:
3𝑥 2 + 3𝑦 2 = 12
3𝑥 2
3
+
3𝑦 2
3
=
12
3
𝒙 𝟐 + 𝒚𝟐 = 𝟒
Now we can see that the centre of the circle is at (0, 0) and the radius is 2 (𝑎𝑠 √4 = 2)
𝑦
2
-2
2
𝑥
-2
© H Jackson 2011 / ACADEMIC SKILLS
2
Ellipse
𝑥2
𝑎2
+
𝑦2
𝑏2
Notice that for an ellipse:
 the 𝑥 and 𝑦 terms are both squared, and then added
 there is no 𝑥𝑦 term.
 the ellipse intercepts the 𝑥-axis at ±𝑎 and the 𝑦-axis at ±𝑏
=1
(n.b. when 𝑎 = 𝑏 we have a circle).
Examples:

Sketch the ellipse with equation
𝒙𝟐
𝟗
𝒚𝟐
+ 𝟐𝟓 = 𝟏
We need squared numbers on the bottom of each fraction so we let’s rewrite the equation as:
𝑥2 𝑦2
+
=1
32 52
It’s now easy to see that the ellipse crosses the 𝑥-axis at ±3 and the 𝑦-axis at ±5
𝑦
5
𝑥
3
−3
−5

Sketch the ellipse with equation 𝟐𝒙𝟐 + 𝟖𝒚𝟐 = 𝟑𝟐
The equation has to equal 1 so we divide the whole equation by 32.
2𝑥 2
32
We now have:
𝒙𝟐
+
+
𝟏𝟔
8𝑦 2
32
𝒚𝟐
𝟒
32
= 32
=𝟏
(we can rewrite this as:
𝑥2
42
𝑦2
+ 22 = 1)
From the equation we see that the ellipse crosses the 𝑥-axis at ±4 and the 𝑦-axis at ±2
𝑦
2
−4
4
𝑥
−2
© H Jackson 2011 / ACADEMIC SKILLS
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Hyperbola
𝑥 2 𝑦2
−
=1
𝑎2 𝑏 2
Notice that for a hyperbola:
 the 𝑥 and 𝑦 terms are both squared and then subtracted.
 the diagonals of the rectangle (formed with length from −𝑎 to 𝑎
and height from −𝑏 to 𝑏) are asymptotes for the hyperbola.
Examples:

Sketch the hyperbola with equation
We can rewrite the equation as:
𝑥2
𝒙𝟐
𝟒
−
𝒚𝟐
𝟗
=𝟏
𝑦2
− 32 = 1
22
From the equation we see that we need to draw a rectangle from ±2 along the 𝑥-axis and
from ±3 along the 𝑦-axis. We then use the diagonals of the rectangle as asymptotes for the
hyperbola. (It might be a good idea to draw the rectangle and it’s diagonals in pencil so you can
erase them when you have used them as a guide to draw your hyperbola).
𝑦
3
2
𝑥
Form a rectangle.

Use the diagonals
as asymptotes.
Sketch the hyperbola with equation 𝟖𝒙𝟐 − 𝟐𝒚𝟐 = 𝟖
The equation has to equal 1 so we need to divide the whole equation by 8.
8𝑥 2
8
We now have:
−
𝑥2 −
2𝑦 2
8
𝑦2
4
8
=8
=1
We can rewrite this as:
𝑥2
12
𝑦2
− 22 = 1
This time draw your rectangle from ±1 along the 𝑥-axis and from ±2 along the 𝑦-axis.
𝑦
2
1
𝑥
© H Jackson 2011 / ACADEMIC SKILLS
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