A hyperbola has vertices at (6, 15) and (6, -5). The perimeter of the rectangle made by the transverse and conjugate axes is 72 units. State the equation.
Both of the vertices have the same x-coordinate, which means that the hyperbola is vertical.
The equation of the (vertical) hyperbola is thus,
(y – k)^2/a^2 – (x – h)^2/b^2 = 1
Centre of hyperbola is (h, k).
Transverse axis is 2a
Conjugate axis is 2b
Rectangle has height H = 2a and width W = 2b
Perimeter is L = 2H + 2W = 4a + 4b
i.e. L = 72 = 4a + 4b
a + b = 18
In a vertical hyperbola, with transverse axis = 2a, the vertices are given by
V1 = (h, k+a) and V2 = (h, k-a)
Comparing V1 and V2 with (6, 15) and V2 = (6, -5), then
h = 6 and
k + a = 15
k – a = -5
From which k = 5, a = 10 and b = 8 (using a + b = 18)
The equation of the hyperbola is thus: (y – 5)^2/10^2 – (x – 6)^2/8^2 = 1