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  • one year ago

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).

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  1. anonymous
    • one year ago
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    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 \[\frac{ y^{2} }{ 4 } - \frac{ x^{2} }{ 117 } = 1\]

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