What is the equation of conjugate hyperbola?

The eccentricity of the hyperbola (i) is e = √a2+b2a2 or, b2 = a2(e2 – 1) and its conjugate hyperbola (ii) is e = √b2+a2b2 or, a2 = b2(e2 – 1).

How do you find the conjugate axis of a hyperbola?

If two points B and B’ are on the y-axis such that CB = CB’ = b, then the line segment BB’ is called the conjugate axis of the hyperbola. Therefore, the length of conjugate axis = 2b. Solved examples to find the transverse and conjugate axes of an hyperbola: 1.

What is conjugate axis and transverse axis?

The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints.

How do you find the equation of a hyperbola?

The equation of the hyperbola is simplest when the centre of the hyperbola is at the origin and the foci are either on the x-axis or on the y-axis. The standard equation of a hyperbola is given as: [(x 2 / a 2) – (y 2 / b 2)] = 1. where , b 2 = a 2 (e 2 – 1)

How do you find the transverse axis of a hyperbola?

Identify the vertices and foci of the hyperbola with equation y2 49 − x2 32 =1 y 2 49 − x 2 32 = 1. The equation has the form y 2 a 2 − x 2 b 2 = 1 y 2 a 2 − x 2 b 2 = 1, so the transverse axis lies on the y -axis. The hyperbola is centered at the origin, so the vertices serve as the y -intercepts of the graph.

What are the conjugate hyperbolas of each other?

2 hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & transverse axis of the other are called conjugate hyperbola of each other. (x 2 / a 2) – (y 2 /b 2) = 1 & (−x 2 / a 2) + (y 2 / b 2) = 1 are conjugate hyperbolas of each other.

How do you find the tangent of a rectangular hyperbola?

The tangent of a rectangular hyperbola is a line that touches a point on the rectangular hyperbola’s curve. The equation and slope form of a rectangular hyperbola’s tangent is given as: The y = mx + c write hyperbola x 2 /a 2 – y 2 /b 2 = 1 will be tangent if c 2 = a 2 /m 2 – b 2.