Find the​ vertices, foci, and eccentricity of the equation.

Question

Find its vertices, foci, and eccentricity: 25(y-4)^2 -16(x-3)^2 =400

Answer 1

25(y-4)² - 16(x-3)² = 400       Divide both sides of the equation by 400.

(y-4)²/16 - (x-3)²/25 = 1        The center is at (3,4).  This is the equation of a hyperbola, which opens up and down.

a² = 16, a = ±4     b² = 25, b = ±5

c² = a² + b²     c² = 4² + 5² = 16 + 25 = 41

c = ±√41 ≈ ±6.4

a = the distance between the center and the vertex.

V(3,8)  V'(3,0)

c = the distance between the center and the focus.

F(3, 10.4)  F'(3, -2.4)

Hyperbolas don't have an eccentricity, only ellipses have one.