8.4 Classifying Conic Section Notes
Part 1) identifying conic sections when they are in standard form
Circle: (x – h)2 + (y – k)2 = r2
Parabola: (x – h)2 = 4p(y – k) or (y – k)2 = 4p(x – h)
Ellipse:
or
Hyperbola:
x  h2   y  k 2
a2
b2
(
 1 or
 y  k 2  x  h2
a2
b2
1
To do: 1) look at the given equation and determine which of the 4 it looks like above
2) write down which 1 it is
Part 2) Identifying the conic section from general form and rewrite in standard form
General form: Ax2 + Cy2 + Dx + Ey + F = 0
You have a:
Cirlce if A = C
Parabola: if AC = 0 but both A and C cannot be 0
Ellipse: if AC > 0
Hyperbola: if AC < 0
To do: 1) determine if you have a circle, parabola, ellipse, or hyperbola by using the information above
2) rewrite so that it’s in the form: Ax2 + Dx + Cy2 + Ey = -F
3) complete the square (before you can do this make sure you have 1x2 or 1y2 – you may have to factor to do so)
Completing the square:
1) Take (D/2)2, add that answer to both sides (if you had to factor out a number multiply (D/2)2 by that number –
you’ll see an example below)
2) Take (E/2)2, add that answer to both sides (if you had to factor out a number multiple (E/2)2 by that number –
you’ll see an example below)
3) Factor both trinomials as perfect square trinomials (square root the first term, sign in front of the 2nd term,
sqaure root the 3rd term, put in a set of parenthesis squared)
4) Add the numbers on the right
4)finish writing the equation in standard form
Examples:
1. 9x2 + 4y2 – 18x + 16y -11 = 0
A = 9, C = 4
9 ≠ 4 so not a circle, 9(4) = 36 ≠ 0 so not a parabola, 9(4) = 36 > 0 so it’s an ellipse
9x2 – 18x + 4y2 + 16y = 11
9(x2 – 2x) + 4(y2 + 4y) = 11 (had to factor out a 9 and a 4 because you must have 1x2 and 1y2)
-2/2 = -12 = 1 4/2 = 22 = 4
9(x2 – 2x + 1) + 4(y2 + 4y + 4) = 11 + 9 + 16 (add 9 because 9(1) = 9, add 16 because 4(4) = 16)
9(x – 1)2 + 4(y + 2)2 = 36 ( to get the equation = 1 divide by 36)
9(x – 1)2 + 4(y + 2)2 = 36
36
36
36
(x – 1)2 + (y + 2)2 = 1
4
9
Final answer:
2. x2 + y2 – 10x + 4y + 27 = 0
A = 1, C = 1
1 = 1 so it’s a circle
x2 – 10x + y2 + 4y = -27
-10/2 = 52 = 25, 4/2 = 22 = 4
x2 – 10x + 25 + y2 + 4y + 4 = -27 + 25 + 4
(x – 5)2 + (y + 2)2 = 2
Final answer: (x – 5)2 + (y + 2)2 = 2
3. 4y2 – 9x2 – 18x – 8y – 41 = 0
-9x2 + 4y2 – 18x – 8y – 41 = 0
A = -9, C = 4
-9≠4 so not a circle, -9(4)= -36 ≠ 0 so not a parabola, -9(4) = -36
0 so not an ellipse, -36 < 0 so hyperbola
-9x2 + 18y + 4y2 – 8y = 41
-9(x2 + 2x) + 4(y2 – 2y) = 41
2/2 = 12 = 1
-2/2 = -12 = 1
2
-9(x + 2x + 1) + 4(y2 – 2y + 1) = 41 – 9 + 4 (don’t forget to do -9(1) and 4(1) since you factored out a -9 and 4)
-9(x + 1)2 + 4(y – 1)2 = 36
-9(x + 1)2 + 4(y – 1)2 = 36
36
36
36
-(x + 1)2 + (y – 1)2 = 1
4
9
Final answer: (y – 1)2 – (x + 1)2 = 1
9
4
4. y2 – 4y – 8x + 20 = 0
A = 0, C = 1
0 ≠ 1 so not a circle, 0(1) = 0 so it’s a parabola
-8x + y2 – 4y = -20
-4/2 = 22 = 4
-8x + y2 – 4y + 4 = -20 + 4
-8x + (y – 2)2 = -16
(y – 2)2 = 8x – 16 (since it’s a parabola only x2 or y2 term can be on the left so add or subtract the x or y term to the
right)
(y – 2)2 = 8(x – 2) (because the equation is (x – h) you must factor out the number in front of x and that will
determine 4p)
Final answer: (y – 2)2 = 8(x – 2)
Part 3) classify the conic section, write the equation in standard form, sketch the graph the find:
Circles: center and radius
Parabola: vertix and focus
Ellipse: center, vertices, and foci
Hyperbola: center, vertices, and foci
Do just like we have been doing