# What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^{2 }- 4y^{2 }= 64.

**Solution:**

The standard equation of hyperbola is x^{2 }/ a^{2} - y^{2 }/ b^{2} = 1 and foci = (± ae, 0) where, e = eccentricity = √[(a^{2} + b^{2}) / a^{2}]. Vertices are (±a, 0) and the equations of asymptotes are (bx - ay) = 0 and (bx + ay) = 0.

Given, 16x^{2} - 4y^{2} = 64

Divide the above equation by 64

x^{2 }/ 4 - y^{2 }/ 16 = 1

Comparing with x^{2 }/ a^{2} - y^{2}/b^{2} = 1

a^{2} = 4, b^{2} = 16 ⇒ a = 2 and b = 4

∴ Vertices = (±a, 0) = (±2, 0)

Eccentricity = e = √[(a^{2} + b^{2}) / a^{2}] = √[(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 16x^{2 }- 4y^{2}= 64.

**Summary:**

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

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