Question

Which standard form of the equation of the hyperbola has vertices at (12, 0) and (-12, 0), and asymptotes y= + 5/12 x?

Answer 1

The standard form of the equation of the hyperbola with given vertices and asymptotes is (x - 12)^2/144 - y^2/b^2 = 1.

Step-by-step explanation:

The standard form of the equation of a hyperbola with vertices at (h, k) and asymptotes y = mx + b is given by:

(x - h)^2/a^2 - (y - k)^2/b^2 = 1.

In this case, the hyperbola has vertices at (12, 0) and (-12, 0), and the asymptotes are y = (5/12)x and y = (-5/12)x.

Since the vertices are at (h, k) = (12, 0), the value of h is 12. The slope of the asymptotes, m, is given by m = ±(b/a), so 5/12 = ±(b/a). We can choose the positive sign since the slope is positive.

Now, we need to find the value of a. We know that a is the distance from the center to each vertex, so a = 12. We can substitute these values into the equation:

(x - 12)^2/12^2 - (y - 0)^2/b^2 = 1.

Simplifying, we get:

(x - 12)^2/144 - y^2/b^2 = 1.

So, the standard form of the equation of the hyperbola is:

(x - 12)^2/144 - y^2/b^2 = 1.

Answer 2

The standard form of the equation of the hyperbola with vertices at (12, 0) and (-12, 0) is x²/144 - y²/25 = 1. The value 'a' equals 12, and 'b' was found using the slope of the given asymptotes y = ± 5/12 x.

Step-by-step explanation:

The standard form of the equation of a hyperbola with vertices at (12, 0) and (-12, 0) is given by x²/a² - y²/b² = 1,
where a is the distance from the center to a vertex along the x-axis and b is related to the asymptotes. Since the vertices are at (12, 0) and (-12, 0), the distance 'a' is 12. Given the asymptotes are y = ± 5/12 x, the slope of the asymptotes 'm' is ± 5/12, which suggests that the relationship b/a = 5/12. Solving for 'b', we get b = 5a/12, substituting 'a' with 12, we get b = 5. Therefore, the equation of the hyperbola is x²/144 - y²/25 = 1.