If a hyperbola passes through the point (4, 0) and has foci at (±5, 0), and the length of its conjugate axis is 6, then the eccentricity is:

The hyperbola has foci at (±5,0), which means the center of the hyperbola is at the origin (0,0) and the major axis lies along the x-axis. The distance from the center to each focus is c. So, c=5. The length of the conjugate axis is 2b=6, which implies b=3. For a hyperbola, the relationship between a, b, and c is c² = a² + b² Substituting the values of b and c:
5² = a² + 3²
a = 4 The equation of the hyperbola with center at the origin and major axis along the x-axis is The point (4,0) indeed lies on the hyperbola. This also means that the vertex of the hyperbola along the positive x-axis is at (4,0), so a=4, which is consistent with our calculation. The eccentricity e of a hyperbola is defined as e=c/a​. Substituting the values of c and a: e = 5/4