Find an equation for the hyperbola that has the vertices 0 1

Solution

We have given the vertices as (0, -12) and (0, 12) We also know that for vertical standard form equation of hyperbola as (y - k)^2/b^2 - (x - h)^2/a^2 = 1 with vertices (h, k + b) and (h, k - b) where (h, k) is the center, a^2 is the distance from the center to the vertices, and b is the distance from the center to the co-vertices. by comparing the given vertices, we will get h = 0 and k + b = -12 and k - b = 12 simplifying this, we will get k = 0 and b = -12 now, plugging the given value will give the hyperbola equation as (y - 0)/(-12)^2 - (x - 0)^2/a^2 = 1 i.e. (y)^2/144 - (x)^2/a^2 = 1 Now, as the passing through the point (-5, 18). We need to plug the passing point values into the hyperbola equation to get the value of a^2. So we have given x = -5 and y = 18 we will get (18)^2/144 - (-5)^2/a^2 = 1 324/144 - 25/a^2 = 1 2.25 - 25/a^2 = 1 -25/a^2 = 1 - 2.25 -25/a^2 = -1.25 -25 = -1.25*a^2 Dividing both the sides by -1.25 will give a^2 = 25/1.25 = 20 So, the required hyperbola equation will be (y)^2/144 - (x)^2/20 = 1 Hope this will help yoU!