The asymptotes of a hyperbola with horizontal transverse axis have equations y=1/3±(x+1)/4 Find the (a) center= (b) vertices= (c) foci= (d) draw the hyperbola

Question

The asymptotes of a hyperbola with horizontal transverse axis have equations y=1/3±(x+1)/4

Find the

(a)   center=

(b)  vertices=

(c)   foci=

(d)  draw the hyperbola

Answer 1

For a left-right opening hyperbola the general form is:

(x-h)²/a² - (y-k)²/b² = 1

where the center of the hyperbola is at (h, k), "a" is the distance from center to a vertex, and "b" is the distance from the center to the conjugate axis. The slopes of the asymptotes are b/a and -b/a. Both asymptotes pass through the center of the hyperbola.

That means in order to write the equations of the asymptotes you can use point-slope. The slope is b/a passing through (h,k): y - k = (b/a)(x - h) or y = k + (b/a)(x - h). Similarly for the asymptote with slope -b/a.

The vertices, center, and foci all lie on the transverse axis. So every one of those points has the same y-coordinate.

Looking at the given problem one of the asymptotes has the equation:

y = 1/3 + (1/4)(x + 1)

So, b = 1, a = 4, h = -1, k = 1/3

Center: (-1. 1/3).

Vertices will be a horizontal distance of 4 from the center:

V₁ = (-1 + 4 , 1/3) = (3, 1/3)

V₂ = (-1 - 4, 1/3) = (-5, 1/3)

"c" is the distance from the center to a focus, and c² = a² + b².

c = √(16 + 1) = √17

F₁ = (-1 + √17, 1/3)

F₂ = (-1 - √17, 1/3)

The equation of the hyperbola:

(x + 1)²/16 - (y - 1/3)²/1 = 1

desmos.com/calculator/oho6xzujjw

To sketch the graph, pinpoint the center, graph the vertices, lightly sketch the conjugate axes (y = 1/3 ± 1), which allows you to locate the corners of the auxiliary rectangle, graph the asymptotes. Finally sketch the hyperbola by starting at a vertex and use the asymptotes to constrain the location of the two branches.