Find the equation of the hyperbola with foci (0,±3)(0, \pm 3)(0,±3) and vertices (0,±112)(0, \pm \frac{\sqrt{11}}{2})(0,±211).
Foci are on y-axis, so equation is y2a2−x2b2=1\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1a2y2−b2x2=1. Given c=3c = 3c=3 and a=112a = \frac{\sqrt{11}}{2}a=211. Relation: c2=a2+b2c^2 = a^2 + b^2c2=a2+b2. 9=114+b29 = \frac{11}{4} + b^29=411+b2 b2=9−2.75=6.25=254b^2 = 9 - 2.75 = 6.25 = \frac{25}{4}b2=9−2.75=6.25=425. Equation: y211/4−x225/4=1\frac{y^2}{11/4} - \frac{x^2}{25/4} = 111/4y2−25/4x2=1 4y2/11−4x2/25=1 ⟹ 100y2−44x2=2754y^2/11 - 4x^2/25 = 1 \implies 100y^2 - 44x^2 = 2754y2/11−4x2/25=1⟹100y2−44x2=275
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