resuelto test de algebra

1. De la siguiente elipse: 3x² + 5y²– 6x - 12 = 0. Determine:

A. Centros B. Focos C. Vértices
3x² + 5y² - 6x -12 = 0
(3x² - 6x) + 5y² - 12 = 0
(x² - 2x /3) + 5/3y² -12/3 = 0
(x² - 2x + 1 /3) + 5/3y² - 12/3 - 1/3 = 0
(x – 1) (x – 1) /3 + 5/3y² - 13/3 = 0
(x – 1)² /3 + 5/3y² = 13/3
(x – 1)² /13 + 5/13y² = 1

Ec general elipse:

(x-h)² / a² - (y – k)² / b² =1

H= 1
K= 0
A= √13 = 3.605
B= √13/5 = 1.612

Centro: (h,k) = (1,0)

Focos: f1 (h – c, k) y f2 (h + c, k)
C= √a² - b²
C= √13 – 13/5 = 3. 225
F1 = (1 – 3.225,0) = f1 ( - 2.225,0)
F2= (1 +3.225,0) = f2 (4.225,0)

Vértices:
A= (h – a, k) A’= (h + a, k)
A= (- 2.605,0) A’= (4.605,0)
B= (h, k + b) B’= (h, k – b)
B= (1, 1.612) B’= (1, - 1.612)

2. De lasiguiente hipérbola: 4y² – 9x² + 16y + 18x = 29. Determine:

A. Centros B. Focos C. Vértices
4y² – 9x² + 16y + 18x = 29
(4y² - 16y) – (9x² - 18x) = 29
(y² - 4y) /4*9 – (x²- 2x) /4*9 = 29 /4*9
(y – 4y + 4) /36 – (x² - 2x + 1) /36 = 29/36 + 4/36 – 1/36
(y – 2 ) /36 – (x – 1)² /36 = 32/36
(y – 2)² /32 – (x – 1)² /32 = 1

Ec general hipérbola:

(y – k)² /a² - (x –h)² /b² = 1

K= 2
H= 1
A= √32 = 5.657
B= √32 = 5.657

Centro: (h, k) = (1, 2)

Focos: f1 (h, k + c) y f2 (h, k - c)
C= √a² + b²
C= √32 + 32 = √54 = 8
F1= (1, 2 + 8) = f1 (1, 10)
F2= (1, 2– 8) = f2 (1, 6)

Vertices:

A= (h, k + a) A’= (h, k - a)
A= (1, 7.657) A’= (1, 3.657)
B= (h + b, k) B’= (h – b, k)
B= (6.657,2) B’= (- 4.657,2)



3. Analicela siguiente ecuación: x² + y² – 6x – 8y + 9 = 0. Determine:

A. Centro B. Radio

X² + y² – 6x – 8y + 9 = 0

(X² - 6x) – (y² - 8y) + 9 = 0
(X² - 6x + 9) (y² - 8y + 16) = + 9 - 9 – 16 =0
(x – 3)² + (y – 4)² = 16

Ecuación General Circunferencia:

(x – h)² + (y – k)² = r²

H= 3
K= 3
R= √16 = 4

Centro: (h, k) = (3, 4)

Radio: r = 4
1. De la siguiente parábola: x² + 6x...