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A hyperbola centered at the origin has verticies at (add or subtract square root of 61,0 and foci at (add or subtract square root of 98,0

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Answer:

[tex]\frac{x^2}{61}-\frac{y^2}{37} =1[/tex]

Step-by-step explanation:

The standard equation of a hyperbola is given by:

[tex]\frac{(x-h)^2}{a^2} -\frac{(y-k)^2}{b^2} =1[/tex]

where (h, k) is the center, the vertex is at (h ± a, k), the foci is at (h ± c, k) and c² = a² + b²

Since the hyperbola is centered at the origin, hence (h, k) = (0, 0)

The vertices is (h ± a, k) = (±√61, 0). Therefore a = √61

The foci is (h ± c, k) = (±√98, 0). Therefore c = √98

Hence:

c² = a² + b²

(√98)² = (√61)² + b²

98 = 61 + b²

b² = 37

b = √37

Hence the equation of the hyperbola is:

[tex]\frac{x^2}{61}-\frac{y^2}{37} =1[/tex]

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