Question

Write the equation in standard form for the hyperbola with vertices (4,0) & (-4,0) and asymptote y=(1/4)x & y=-(1/4)x.

Please help me and explain. Thank you.

Answer 1

The graph of the equation on the left has the following properties: x intercepts at ± a , no y intercepts, foci at (-c , 0) and (c , 0), asymptotes with equations y = ± x (b/a)

The hyperbola standard form is x²/a² + y²/b² = 1------>(1)

Given that vertices (4,0) & (-4,0) and asymptote y=(1/4)x & y=-(1/4)x.

asymptotes with equations y = ± x (b/a) = ± x (1/4)

So, a = 4 and b = 1.

Substitute a = 4 and b = 1 in the equation(1).

[( x²/4²) + (y²/1²)] =1

[( x²/16) + (y²/1)] =1

The hyperbola standard form is x²/16 + y²/1 = 1

Answer 2

The vertices of hyperbola is (4, 0) and (- 4, 0) and asymptote y = ± (1/4)x.

The standard form of equation of hyperbola with center at the origin (where a and b are not equals to 0) is x ²/a ² - y ²/b ² = 1 (Transverse axis is horizontal) or y ²/a ²- x ²/b ² = 1 (Transverse axis is vertical).

The x - coordinates of the  vertices points are 4 and - 4.

The value of a = 4, because the vertices are four units from the center.

Because the transverse axis is horizontal, the asymptotes are of the forms y = (b/a) x and y = - (b/a) x.

The asymptote y = ± (1/4)x is comparison with y = ± (b/a) x.

b/a = 1/4 ------> b = 1 and a = 4.

The standard form of the hyperbola equation is x ²/4 ² - y ²/1 ² = 1.