Identify the coordinates of
with the given equation. Then find the length of latus rectum and graph the parabola.
Problem 1 :
y = (x - 3)2 - 4
Problem 2 :
y = (x - 4)2 + 3
Problem 3 :
x = (-1/3) y2 + 1
Problem 4 :
x = 3(y + 1)2 - 3
Problem 5 :
x = y2 - 14y + 25
1)
Direction of opening |
Opening up |
Vertex |
(h, k) ==> (3, -4) |
Focus |
(h, k + a) ==> (3, -4 + 1/4) (3, -15/4) |
Equation of latus rectum |
y = k + a y = -4 + 1/4 = -15/4 |
Equation of directrix |
y = k - a y = -4 - (1/4) y = -17/4 |
Equation of axis of symmetry |
x = h x = 3 |
Length of latus rectum |
4a = 1 1 unit |
2)
Direction of opening |
Opening up |
Vertex |
(h, k) ==> (4, 3) |
Focus |
(h, k + a) ==> (4, 3 + 1/4) (3, 13/4) |
Equation of latus rectum |
y = k + a y = 3 + 1/4 = 13/4 |
Equation of directrix |
y = k - a y = 3 - (1/4) y = 11/4 |
Equation of axis of symmetry |
x = h x = 4 |
Length of latus rectum |
4a = 1 unit |
3)
Direction of opening |
Opening up |
Vertex |
(h, k) ==> (1, 0) |
Focus |
h - a = 1 - (3/4) = 1 - 3/4 = 1/4 (h - a, k) ==> (1/4, 0) |
Equation of latus rectum |
x = h - a x = 1 - 3/4 = 1/4 |
Equation of directrix |
x = h + a x = 1 + (3/4) x = 7/4 |
Equation of axis of symmetry |
y = k y = 0 |
Length of latus rectum |
4a = 3 units |
4)
Direction of opening |
Opening up |
Vertex |
(h, k) ==> (-3, -1) |
Focus |
h + a = -3 + (1/12) = -35/12 (h + a, k) ==> (-35/12, -1) |
Equation of latus rectum |
x = h + a x = -35/12 |
Equation of directrix |
x = h - a x = -3 - (1/12) x = -37/12 |
Equation of axis of symmetry |
y = k y = -1 |
Length of latus rectum |
4a = 1/3 units |
5)
Direction of opening |
Opening up |
Vertex |
(h, k) ==> (-24, 7) |
Focus |
h + a = -24 + (1/4) = -95/4 (h + a, k) ==> (-95/4, 7) |
Equation of latus rectum |
x = h + a x = -95/4 |
Equation of directrix |
x = h - a x = -24 - (1/4) x = -97/4 |
Equation of axis of symmetry |
y = k y = 7 |
Length of latus rectum |
4a = 1 unit |
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM