Use the vertices and asymptotes to graph each hyperbola. Locate the foci and find the equation of asymptotes.
Problem 1 :
Problem 2 :
Problem 3 :
Problem 4 :
4y2 - x2 = 1
Problem 5 :
(x - 3)2 - 4(y + 3)2 = 4
Problem 6 :
1)
Center : C (0, 0)
Vertices : A(3, 0) and A'(-3, 0) and c = 5
Foci : F1 (5, 0) and F2 (-5, 0)
Equation of asymptotes : y = (-4/3) and y = (4/3) x
2)
Center : C (0, 0)
Vertices : A(10, 0) and A'(-10, 0) and c = 2√41
Foci : F1 (2√41, 0) and F2 (-2√41, 0)
Equation of asymptotes : y = (-4/5) and y = (4/5) x
3)
Center : C (0, 0)
Vertices : A(0, 4) and A'(0, -4) and c = 2√13
Foci : F1 (0, 2√13) and F2 (0, 2√13)
Equation of asymptotes : y = (-2/3)x and y = (2/3) x
4)
Center : C (0, 0)
Vertices : A (0, 1/2) and A'(0, -1/2) and c = √5/2
Foci : F1 (0, √5/2) and F2 (0, -√5/2)
Equation of asymptotes : y = (-1/2)x and y = (1/2) x
5)
Center : C (h, k) ==> (3, -3)
Vertices : A(5, -3), A'(1, -3) and c = √5
Foci : F1 (3 + √5 , -3) and F2 (3 - √5 , -3)
Equation of asymptotes :
y = (1/2)x - (9/2) and y = (-1/2)x + (-3/2)
6)
Center : C (h, k) ==> (1, -2)
Vertices : A(1, 0), A'(1, -4) and c = 2√5
Foci :F1 (1, -2 + 2√5 ) and F2 (1, -2 - 2√5)
Equation of asymptotes :
y = (1/2)x - (5/2) and y = (-1/2)x - (3/2)
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM