What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x² - 4y² = 64.

Solution:

The standard equation of hyperbola is x² / a² - y² / b² = 1 and foci = (± ae, 0) where, e = eccentricity = √[(a² + b²) / a²]. Vertices are (±a, 0) and the equations of asymptotes are (bx - ay) = 0 and (bx + ay) = 0.

Given, 16x² - 4y² = 64

Divide the above equation by 64

x² / 4 - y² / 16 = 1

Comparing with x² / a² - y²/b² = 1

a² = 4, b² = 16 ⇒ a = 2 and b = 4

∴ Vertices = (±a, 0) = (±2, 0)

Eccentricity = e = √[(a² + b²) / a²] = √[(4 + 16) / 4] = √5

Foci = (±ae, 0) = [±2√5, 0]

Asymptotes as (4x - 2y) = 0 and (4x + 2y) = 0

⇒ (2x - y) =0 and (2x + y) =0

What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x² - 4y²= 64.

Summary:

The vertices, foci and asymptotes of the hyperbola with the equation 16x² - 4y² = 64 are (±2, 0), [±2√5, 0], (2x - y) = 0 and (2x + y) = 0 respectively.