Answer:
The order is 4) → 5) → 6) → 7) → 2) → 1) → 3)
Please find diagram with the arrangements
Step-by-step explanation:
The horizontal width of an hyperbola
For
1) [tex]\dfrac{(y - 11)^2}{7^2} -\dfrac{(x - 2)^2}{6^2} = 1[/tex]
h = 2, k = 11
The widths are;
Horizontal (h - a, k) to (h + a, k) which is (2 - 7, 11) to (2 + 7, 11) = 14 units wide
(h, k - b) to (h, k + b) which is (2, 11 -6) to (2, 11 + 6) = 12 units wide
2)
[tex]\dfrac{(y - 1)^2}{5^2} -\dfrac{(x - 7)^2}{12^2} = 1[/tex]
h = 7, k = 1
(h - a, k) to (h + a, k) which is (7 - 5, 1) to (7 + 5, 1) = Horizontal width 10 units wide
(h, k - b) to (h, k + b) which is (7, 1 -12) to (7, 1 + 12) = 24 units wide
3) [tex]\dfrac{(x - 6)^2}{6^2} -\dfrac{(y + 1)^2}{3^2} = 1[/tex]
h = 6, k = -1
a = 8, b = 3
The widths are;
(6 - 8, -1) to (6 + 8, -1) Horizontal width = 16
(6, -1 - 3) to 6, -1 + 3) width = 6
4) [tex]\dfrac{(x - 4)^2}{2^2} -\dfrac{(y + 2)^2}{5^2} = 1[/tex]
h = 4, k = -2, a = 2, b = 5
(4 - 2, (-2)) to (4 + 2, (-2)) Horizontal width = 4
(4, -2 - 5) to (4, -2 + 5) width = 10
5) [tex]\dfrac{(y + 5)^2}{2^2} -\dfrac{(x + 4)^2}{3^2} = 1[/tex]
h = -4, k = -5, a = 2, b = 3
(-4 - 2, (-5)) to (-4 + 2, (-5)) Horizontal width = 4
(-4, -5 - 3) to (4, -5 + 3) width = 6
6) [tex]\dfrac{(y + 1)^2}{2^2} -\dfrac{(x - 1)^2}{9^2} = 1[/tex]
h = 1, k = -1, a = 2, b = 9
(1 - 2, (-1)) to (1 + 2, (-1)) Horizontal width = 4
(1, -1 - 9) to (1, -1 + 9) width = 18
7) [tex]\dfrac{(x + 7)^2}{4^2} -\dfrac{(y - 9)^2}{9^2} = 1[/tex]
h = -7, k = 9, a = 4, b = 9
(-7 - 4, 9) to (-7 + 4, 9) Horizontal width = 8
(-7, 9 -9) to (-7, 9 + 9) width = 18
