Conics
Conics: 2nd degree (non-linear equations)


Parabola

Circle

Ellipse
Hyperbola
Parabola



Locus of a point that moves so that its distance from a fixed point (focus) is always equal to its
distance from a fixed straight line (directrix).
Axis of Symmetry: line through the focus and perpendicular to the directrix.
When looking to graph parabolas, need to find the following information
o Axis of symmetry
o Vertex
2
 Y = ax (opens up/down)
 X = ay2 (opens right/left)
Focus & Directrix for y = ax2
if a > 0 then
Focus & Directrix for x = ay2
if a < 0 then
 Parabola opens upward
 Focus is at (0,c)
 Directrix is y = -c
Note: |a| =
if a > 0 then
 Parabola opens downward
 Focus is at (0,-c)
 Directrix is y = c



if a < 0 then
Parabola opens right
Focus is at (c,0)
Directrix is x = -c



Parabola opens left
Focus is at (-c,0)
Directrix is x = c
𝟏
𝟒𝒄
Example: Identify Focus and Directrix
1
1. y = − 16 𝑥 2
Step 1: identify type of parabola
Since in form of y = ax2,
vertical A.O.S
Since a < 0, opens downward
Focus (0, -c)
Directrix y = c
Step 2: Solve for c
1
|a| = 4𝑐
−1
Step 3: Write focus & directrix
Focus (0, -4)
1
| |=
16
4𝑐
4c = 16
c=4
Directrix y = 4
Practice: Identify Focus and Directrix
1
2. y = 12 𝑥 2
1
1
3. x = -8 𝑦 2
4. x = 2 𝑦 2
Focus:
Focus:
Focus:
Directrix:
Directrix:
Directrix:
Example: Writing equation of parabola with vertex at origin
5. Focus at (-5, 0)
Step 1: Determine orientation
Since focus is (-c, 0)
Open left
x = ay2
Step 2: Solve for a
1
|a| = 4(5)
|𝑎| =
Step 3: Write equation
1
X = − 20 𝑦 2
1
20
1
a = − 20
Practice: Writing equation of parabola with vertex at orign
6. Focus at (1/2, 0)
7. Directrix x = 2
1
8. Directrix y = 4
Rev C
Conics
Translating Parabolas
Vertex (h,k)
Y – k = a(x-h)2 or y = a(x-h)2 + k
x – h = a(y-k)2 or x = a(y-k)2 + h
Vertex (0,0)
Y = ax2
X = ay2
Example: Graphing Equation of Parabola
Identify the vertex, focus and directrix and then graph.
9. y2 – 4x- 4y + 16 =0
y
Solve for x, since x is in only 1 term
4x = y2 – 4y + 16
5
4
3
2
Complete the square in y. Get in vertex form.
4x = (y2 – 4y + 4) + 16 – 4
1
X = 4 (𝑦 − 2)2 + 3
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
1
2
3
4
5
x
1
2
3
4
5
x
–2
–3
Vertex: (3, 2)
–4
–5
Find c.
1
|a| = 4𝑐
1
Focus: (4, 2)
Directrix: x = 2
1
|4| = 4𝑐
4c = 4 c = 1
Practice: Graphing Equation of Parabola
Identify the vertex, aos, focus and directrix and then graph.
10. y = (x – 4)2 + 7
y
Vertex:
AOS:
5
4
3
2
1
Focus:
–5
Directrix:
–4
–3
–2
–1
–1
–2
–3
–4
–5
11. x – 4 = 2(y + 3)2
y
5
Vertex:
4
AOS:
3
2
1
–5
Focus:
Directrix:
–4
–3
–2
–1
–1
–2
–3
–4
–5
2
Rev C
Conics
Circle

Locus of a point that moves in such a way that its distance from a fixed point(center) is always
constant (radius)



R = ( x  h) 2 ( y  k ) 2 or (x – h)2 + (y – k) 2 = r2 where center = (h, k)
If center is (0,0) → x2 + y2 = r2
When looking to graph circles, need to find the following information
 Radius
 Vertices (x and y-intercepts if center is (0,0))
 Center
Examples: Writing Equations of Circle
Write an equation for each circle.
1. Center at (-2, 5), d = 4
Step 1: Identify h & k
h = -2, k = 5
2. Center at origin, r = 3
Step 1: Identify h & k
h = 0, k = 0
3. Center at (-12, -1), r = 8
Step 1: Identify h & k
h = -12, k = -1
Step 2: Identify r
d = 4 → r = 4/2 = 2
Step 2: Identify r
r=3
Step 2: Identify r
r=8
Step 3: Substitute in
(x-(-2))2 + (y-5)2 = 22
(x+ 2)2 + (y-5)2 = 4
Step 3: Substitute in
(x - 0)2 + (y-0)2 = 32
x2 + y2 = 9
Step 3: Substitute in
(x – (-12))2 + (y-(-1))2 = 82
(x+ 12)2 + (y+1)2 = 64
Practice: Writing Equations of Circle
Write an equation for each circle.
4. Center at (-2, 5), d = 20
5. Center at origin, r = √7
6. Diameter with endpoints at
(2, 7) and (-6, 15)
Examples: Graphing Equations of Circle
7. (x+ 2)2 + (y-3)2 = 16
Center:
y
5
4
3
Radius:
2
1
–5
Vertices:
–4
–3
–2
–1
–1
1
2
3
4
5
x
1
2
3
4
5
x
–2
–3
–4
–5
8. x2 + y2 = 25
Center:
y
5
4
3
Radius:
2
1
Vertices:
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
3
Rev C
Conics
Ellipse







Locus of a point that moves so that the sum of its distances from 2 fixed points (foci) is constant.
Center: midpoint of segment joining foci
Major Axis: Longest diagonal. Contains the foci and has its endpoints on the ellipse.

Length determined by 2a (if a > b) or 2b (if b > a)

Endpoints of major axis are vertices of ellipse

Midpoint of major axis is center of ellipse
Minor Axis: shortest diagonal.
 Length determined by 2a( if a < b) or 2b (if b < a)
 Perpendicular to major axis at center
 Endpoints of minor axis are co-vertices
x2 y2
( x  h) 2 ( y  k ) 2
Equation with Horizontal Major Axis: 2  2  1 or

 1 Foci (±c, 0)
a2
b2
a b
𝑥2
𝑦2
(𝑥−ℎ)2
(𝑦−𝑘)2
Equation with Vertical Major Axis: 𝑏2 + 𝑎2 = 1or 𝑏2 + 𝑎2 = 1 Foci (0, ±c)
When looking to graph ellipses, need to find the following information

Center

Major/Minor length

Vertices (x and y-intercepts if center is (0,0))
Examples: Writing Equation of Ellipse
1. Write an equation in standard form of an ellipse that has a vertex at (0,5), a co-vertex at (2,0) and a
center at the origin.
Step 1: identify major axis
Step 3: Select correct equation
𝑥2
𝑦2
(0,5) is vertex, therefore vertical
Since vertical is major, use 𝑏2 + 𝑎2 = 1
Vertices: (0, 5) and (0, -5) and a = 5
𝑥2
𝑦2
+
=1
4
25
Step 2: identify minor axis
(2, 0) is co-vertex, therefore horizontal
Co-vertices: (2, 0) and (-2, 0) and b = 2
2. Find an equation of an ellipse centered at the origin that is 20 units wide and 10 units high.
Step 1: identify major axis
Step 3: Select correct equation
𝑥2
𝑦2
Since wider than higher, horizontal
Since horizontal is major, use 𝑎 + 𝑏2 = 1
a = ½(20) = 10
𝑥2
𝑦2
+
=1
100 25
Step 2: identify minor axis
b = ½ (10) = 5
Practice: Writing Equation of Ellipse
3. Write an equation in standard form for an ellipse that has a vertex at (0, -6), a co-vertex at (3, 0) and
a center at the origin.
4. Find an equation of an ellipse centered at the origin that is 12 units wide and 30 units high.
4
Rev C
Conics
Examples: Finding Foci/Using of Ellipse
5. Find the foci of the ellipse with the equation 25x2 +9y2 = 225
Step 1: Write in standard form.
Step 3: Find c (c2 = a2-b2)
2
2
𝑥
𝑦
C=4
+
=1
9
25
Step 4: Write foci, vertices & co-vertices
Foci: (0, ±4)
Vertices: (0, ±5)
Co-vertices: (±3, 0)
Step 2: Identify major axis
Since 25 > 9 and 25 is with y2,
major axis is vertical
a = 5 and b = 3
6. Write an equation of ellipse with foci at (±7,0) and co-vertices at (0, ±6).
Step 1: Identify major axis.
Step 2: Find a2
Step 3: Write equation
2
2
2
𝑥2
𝑦2
Since foci at (±7,0),
C =a –b
+
=1
85
36
Major axis horizontal
a2 = 85
b=6
Practice: Finding Foci/Using of Ellipse
7. Find the foci of the ellipse with the equation 9x2 +y2 = 36
8. Write an equation of the ellipse with foci at (0, ±√17) and co-vertices at (±8,0).
Example: Graph Ellipse
9. Find the center, vertices and major/minor axes of the ellipse
x2 y2

 1 and then graph.
9 25
y
Center:
X-intercepts:
y-Intercepts:
5
4
3
2
Major Axis:
Major Length:
Vertices:
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
–2
–3
Minor Axis:
Minor Length:
Co-Vertices:
–4
–5
5
Rev C
Conics
Hyperbola

Locus of a point that moves so that the difference of its distances from 2 fixed points (foci) is constant.


Center: midpoint of segment joining foci
Foci: (±c, 0) (c2 = a2 + b2)
x2 y2
( x  h) 2 ( y  k ) 2
Open Right/Left: 2  2  1 or

1
a
b
a2
b2
 Horizontal Transverse Axis
 Vertices (±a, 0)
 X-Intercepts: ±a
 Y-Intercepts: none
b
 Asymptotes: y =  x
a
2
2
2
y
x
(y  k)
( x  h) 2

1
Open Up/Down 2  2  1 or
b
a
b2
a2
Vertical Transverse Axis
 Vertical Transverse Axis
 Vertices (0, ±b)
 X-Intercepts: none
 Y-Intercepts: ±b
b
 Asymptotes: y =  x
a
When looking to graph hyperbolas, need to find the following information





Center

Asymptotes
Vertices (x-intercept when x 1st
y-intercepts when y 1st)
Hyperbola Example:
1. Find the foci, vertices and asymptotes of the hyperbola 25x2 - 4y2 = 400.
Step 1: Write in standard form
Step 3: Identify the vertices
𝑥2
𝑦2
(±4, 0)
−
=1
16
100
Step 5: Identify asymptotes: y = ±
Step 2: Identify Transverse Axis, a, b and c
Horizontal, a = 4, b = 10 c = √116
9x2 – 4y2 = 36
Focus:
Vertices:
Asymptotes:
2.
3.
𝑥2
9
𝑦2
y
5
5
4
4
4
3
3
–3
–2
3
2
2
1
1
–1
–1
4
y
5
–4
10
4. 16x2 – y2 = 64
Focus:
Vertices:
Asymptotes:
−
=1
16
Focus:
Vertices:
Asymptotes:
y
–5
Step 4: Identify foci:
(±√116, 0)
1
2
3
4
5
x
–5
–4
–3
–2
–1
–1
2
1
1
2
3
4
5
x
–5
–4
–3
–2
–1
–1
–2
–2
–2
–3
–3
–3
–4
–4
–4
–5
–5
–5
6
1
2
3
4
Rev C
5
x
Conics
7
Rev C