ALGEBRA UNAD

Aporte Individual

1. De la siguiente elipse: 4x2 + 16y2 – 8x – 96y + 84= 0. Determine:

• Centro

• Focos

• Vértices

Solución:

Todo es elevado a la 2

4x2 + 16y2 - 8x - 96y + 84 = 0

4(x2 - 2x) + 16(y2- 6) + 84 = 0

4[(x-1)2 - 1] + 16[(y-3)2 - 9] + 84 = 0

4(x-1)2 - 4 + 16(y-3)2 - 144 + 84 = 0

4(x-1)2 + 16(y-3)2 - 64 = 0

4(x-1)2 + 16(y-3)2 = 64

(x-1)2 + 4(y-3)2 = 16

(x−1)2 + (y-3)2 = 1

16 4

(x−1)2 + (y-3)2 =1

42 22

Centro: (1,3)

Focos: (1,3) – (2√3,0) = (1 - 2√3,3)

(1,3) + (2√3,0) = (1 + 2√3,3)

Vértices:

V1 = (1,3) – (4, 0) = (-3,3)

V2 = (1,3) + (4,0) = (5,3)

V3 = (1,3) – (0,2) = (1, 1)

V4 = (1,3) + (0,2) = (1,5)

2. De la siguiente hipérbola: 4x2 – y2 – 8x – 4y – 4 = 0. Determine

• Centro

• Focos

• Vértices

Solución:

Todo es elevado a la 2

4x – – 8x – 4y – 4 = 0

4 – 8x – + 4y) – 4 = 0

4 – 8x – - 4y = 4

4( – 8x) – ( + 4y)= 4

4( – 2x) – 1( + 4y) = 4

4( – 2x+ 1) -1( +4y+4 = 4+4-4

4(x-1)

Centro: (1,-2)

Focos:

F1 = (1,-4) – (√5,0) = (1 - √5,-4)

F2 = (1,-4) +

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