Conic Sections “By Definition” Pre-calculus

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

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

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+1 2 + 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

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

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

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

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+2 2 + 𝑦−2 2 + (𝑥+2) 2 + (𝑦−2) 2 𝑥 2 −12𝑥+36 =100−20 𝑥+2 2 + 𝑦−2 2 + 𝑥 2 +4𝑥+4 20 (𝑥+2) 2 + (𝑦−2) 2 = 16𝑥+68 ( ) 2 ( ) 2

Conic Section 3600 3600 #4 Cont’d 400 𝑥+2 2 + 𝑦−2 2 400 𝑥+2 2 + 𝑦−2 2 =256 𝑥 2 +2176𝑥+4624 400 𝑥 2 +4𝑥+4+ 𝑦 2 −4𝑦+4 =256 𝑥 2 +2176𝑥+4624 400 𝑥 2 +1600𝑥+400 𝑦 2 −1600𝑦+3200 =256 𝑥 2 +2176𝑥+4624 144 𝑥 2 −576𝑥+400 𝑦 2 −1600𝑦=1424 144 𝑥 2 −4𝑥+ +400 𝑦 2 −4𝑦+ =1424+ + 4 4 576 1600 144 (𝑥−2) 2 +400 (𝑦−2) 2 =3600 3600 3600 (𝑥−2) 2 25 + (𝑦−2) 2 9 =1

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

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 =4+4 𝑥+2 2 + 𝑦−1 2 + (𝑥+2) 2 + (𝑦−1) 2 𝑥 2 −8𝑥+16 =4+4 𝑥+2 2 + 𝑦−1 2 + 𝑥 2 +4𝑥+4 ( ) 2 −12x+8 = 4 (𝑥+2) 2 + (𝑦−1) 2 ( ) 2

Conic Section 128 128 #5 Cont’d 144 𝑥 2 −192𝑥+64 144 𝑥 2 −192𝑥+64=16 𝑥+2 2 + 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+ =16+ + 1 1 128 −16 128 (𝑥−1) 2 −16 (𝑦−1) 2 =128 128 128 (𝑥−1) 2 1 − (𝑦−1) 2 8 =1

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

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

“Conics by Definition” Worksheet Assignment “Conics by Definition” Worksheet