1. Conic Sections “By Definition”
Pre-calculus

2. Learning Objective To find the equation in general form
To determine conics and its equation by definitions

3. When finding the equation given a description, draw a sketch using an arbitrary point 𝑃(𝑥, 𝑦)
Conic Section
You will need the distance formula:
𝑑= ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2

4. 1. For each point, its distance from the fixed point (2, 1) is three times its distance from the fixed point (–1, 4)
Conic Section
P(x, y)
𝑑 2 is 3 times 𝑑 1 d1 d2
𝑑 2 =3 𝑑 1
(–1, 4)
(2, 1)
( ) 2
(𝑥−2) 2 + (𝑦−1) 2
=3 (𝑥+1) 2 + (𝑦−4) 2
( ) 2
(𝑥−2) 2 + (𝑦−1) 2 =9 𝑥 𝑦−4 2
𝑥 2 −4𝑥+4+ 𝑦 2 −2𝑦+1
=9( 𝑥 2 +2𝑥+1+ 𝑦 2 −8𝑦+16)
𝑥 2 + 𝑦 2 −4𝑥−2𝑦+5
=9 𝑥 2 +9 𝑦 2 +18𝑥−72𝑦+153
0=8 𝑥 2 +8 𝑦 2 +22𝑥−70𝑦+148
Since A = C 
Circle

5. 2. Each point is equidistant from the point (–2, 3) and the line 𝑦=−2
Conic Section
(–2, 3)
P(x, y) d1 d2
y = –2
(x, –2)
𝑑 1 = 𝑑 2
( ) 2
(𝑥+2) 2 + (𝑦−3) 2
= (𝑥−𝑥) 2 + (𝑦+2) 2
( ) 2
(𝑥+2) 2 + (𝑦−3) 2 = (𝑦+2) 2
𝑥 2 +4𝑥+4+ 𝑦 2 −6𝑦+9
= 𝑦 2 +4𝑦+4
𝑥 2 +4𝑥−10𝑦+9=0
Only 1 squared term 
Parabola

6. 3. For each point, its distance from the line
3. For each point, its distance from the line 𝑦=−2 is twice its distance from the line 𝑥=4
Conic Section
x = 4
P(x, y) d2
(4, y) d1
𝑑 1 is twice 𝑑 2
y = –2
𝑑 1 =2 𝑑 2
(x, –2)
( ) 2
(𝑥−𝑥) 2 + (𝑦+2) 2
=2 (𝑥−4) 2 + (𝑦−𝑦) 2
( ) 2
(𝑦+2) 2 = 4(𝑥−4) 2
𝑦 2 +4𝑦+4
=4 𝑥 2 −8𝑥+16
𝑦 2 +4𝑦+4=4 𝑥 2 −32𝑥+64
0=4 𝑥 2 − 𝑦 2 −32𝑥−4𝑦+60
Hyperbola

7. We can also use the definitions of each conic section to find their equations. Let’s put final answers in standard form.
Conic Section
Ellipse:
The set of all points P in a plane such that the sum of the distance from P to two fixed points (focus/foci) is constant. P d1 d2 F F
Aka major axis
𝑑 1 + 𝑑 2 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡=2𝑎

8. 4. Determine the equation of an ellipse with foci (6, 2) & (–2, 2) and major axis of length 10.
Conic Section
P(x, y) d1
𝑑 1 + 𝑑 2 =10 d2
(–2, 2)
(6, 2)
𝑑 2 =10− 𝑑 1
( ) 2
(𝑥−6) 2 + (𝑦−2) 2
=10− (𝑥+2) 2 + (𝑦−2) 2
( ) 2
(𝑥−6) 2 + (𝑦−2) 2
=100−20 𝑥 𝑦− (𝑥+2) 2 + (𝑦−2) 2
𝑥 2 −12𝑥+36
=100−20 𝑥 𝑦− 𝑥 2 +4𝑥+4
20 (𝑥+2) 2 + (𝑦−2) = 16𝑥+68
( ) 2
( ) 2

9. Conic Section 3600 3600 #4 Cont’d 400 𝑥+2 2 + 𝑦−2 2
400 𝑥 𝑦−2 2
=256 𝑥 𝑥+4624
400 𝑥 2 +4𝑥+4+ 𝑦 2 −4𝑦+4
=256 𝑥 𝑥+4624
400 𝑥 𝑥+400 𝑦 2 −1600𝑦+3200
=256 𝑥 𝑥+4624
144 𝑥 2 −576𝑥+400 𝑦 2 −1600𝑦=1424
144 𝑥 2 −4𝑥 𝑦 2 −4𝑦+ = 4 4
576
1600
144 (𝑥−2) (𝑦−2) 2 =3600
(𝑥−2) (𝑦−2) 2 9 =1

10. Hyperbola:
Conic Section
P(x, y) d1 d2 F F
The set of all points P in a plane such that the absolute value of the difference of the distance from two fixed points (foci) is a constant.
𝑑 1 − 𝑑 2 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡=2𝑎
𝑑 1 =2𝑎+ 𝑑 2
aka transverse axis

11. 5. Determine the equation of the hyperbola with foci (4, 1) & (–2, 1) with transverse axis of length 2.
Conic Section
P(x, y) d1
𝑑 2 − 𝑑 1 =2 d2
(–2, 1)
(4, 1)
𝑑 2 =2+ 𝑑 1
( ) 2
(𝑥−4) 2 + (𝑦−1) 2
=2+ (𝑥+2) 2 + (𝑦−1) 2
( ) 2
(𝑥−4) 2 + (𝑦−1) 2
= 𝑥 𝑦− (𝑥+2) 2 + (𝑦−1) 2
𝑥 2 −8𝑥+16
= 𝑥 𝑦− 𝑥 2 +4𝑥+4
( ) 2
−12x+8 = (𝑥+2) 2 + (𝑦−1) 2
( ) 2

12. Conic Section 128 128 #5 Cont’d 144 𝑥 2 −192𝑥+64
144 𝑥 2 −192𝑥+64=16 𝑥 𝑦−1 2
144 𝑥 2 −192𝑥+64
=16 𝑥 2 +4𝑥+4+ 𝑦 2 −2𝑦+1
144 𝑥 2 −192𝑥+64
=16 𝑥 2 +64𝑥+16 𝑦 2 −32𝑦+80
128 𝑥 2 −256𝑥−16 𝑦 2 +32𝑦=16
128 𝑥 2 −2𝑥 −16 𝑦 2 −2𝑦+ = 1 1
128
−16
128 (𝑥−1) 2 −16 (𝑦−1) 2 =128
(𝑥−1) 2 1 − (𝑦−1) 2 8 =1

13. Parabola:
Conic Section F d1
P(x, y)
𝑑 1 = 𝑑 2 d2
The set of all points P in a plane that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line.
6. Determine the equation of the parabola with focus (6, –10) and directrix 𝑥=−2 d2
(–2, y)
P(x, y) d1
(6, –10)
x = –2
d1 = d2

14. Conic Section #6 Cont’d ( ) 2 ( ) 2 (𝑥−6) 2 + (𝑦+10) 2
( ) 2
(𝑥−6) 2 + (𝑦+10) 2
= (𝑥+2) 2 + (𝑦−𝑦) 2
( ) 2
(𝑥−6) 2 + (𝑦+10) 2
= (𝑥+2) 2
𝑥 2 −12𝑥+36+ 𝑦 2 +20𝑦+100= 𝑥 2 +4𝑥+4
𝑦 2 +20𝑦=16𝑥−132
𝑦 2 +20𝑦 =16𝑥−132+
100
100
(𝑦+10) 2 =16𝑥−32
(𝑦+10) 2 =16(𝑥−2)
1 16 (𝑦+10) 2 =𝑥−2

15. “Conics by Definition” Worksheet
Assignment
“Conics by Definition” Worksheet