anonymous
  • anonymous
Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Notice that the vertices and foci all have the same x-coordinate. This means that our hyperbola is opening up and down and we have a hyperbola of this form \[\frac{ (y-k)^{2} }{ a^{2} }- \frac{ (x-h)^{2} }{ b^{2} } = 1\] The a^2 always goes with the fraction that is positive and the hyperbola opens along the axis of the variable which is positive in the equation. So, the vertices and foci are all equadistant from the center, meaning the center must be at (0,0), since one vertex is 2 above the origin and one is 2 below. So that let's us know we have this equation now: \[\frac{ y^{2} }{ a^{2} } - \frac{ x^{2} }{ b^{2} } = 1\] The letter that is underneath the x-value represents a distance left and right from the center while the number underneath the y-value represents a distance up and down from the center. Since we have vertices 2 above and below the center, we hve an a value of 2. This means a^2 is 4. In a hyperbola, the foci are determined to be a distance of c units from the center where c is determined by the equation \(a^{2} + b^{2} = c^{2}\). Since the foci are 11 units away from the center, we can plug in a^2 = 4 and c^2 = 11^2 = 121 and solve for b^2. \(a^{2} +b^{2} = c^{2}\) \(4 + b^{2} = 121\) \(b^{2} = 117\) Therefore the equation we're looking for is \[\frac{ y^{2} }{ 4 } - \frac{ x^{2} }{ 117 } = 1\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.