SOLUTION: what would be the vertices, foci, and asymptotes of x^2/9-y^2/36=1

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Question 385710: what would be the vertices, foci, and asymptotes of x^2/9-y^2/36=1
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The equation represents a hyperbola. The center is at (0,0). Here a%5E2+=+9,b%5E2+=+36, b%5E2+=+c%5E2+-+a%5E2, or c%5E2+=+a%5E2+%2B+b%5E2+=+9%2B36+=+45. Hence the foci are (-3sqrt%285%29, 0) and (3sqrt%285%29, 0). Vertices are (-3,0) and (3,0). We can get the asymptotes by considering +x%5E2%2F9+=+y%5E2%2F36. This equation is equivalent to the equations y = -2x and y = 2x, and are the asymptotes of the hyperbola.