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Find the center, vertices, foc, and asymptot ((x-1)²)/(49)-((y-9)²)/(25)=1

User Minillinim
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Final answer:

The center of the hyperbola is (1,9), the vertices are (-6,9) and (8,9), the foci are at (1±√74,9), and the equations of the asymptotes are y-9 = ±(5/7)(x-1).

Step-by-step explanation:

The equation ((x-1)²)/(49)-((y-9)²)/(25)=1 represents a hyperbola. To find its center, vertices, foci (often written as 'foc'), and asymptotes, we should recognize the equation is in the standard form A(x-h)²/B - C(y-k)²/D = 1, where (h,k) is the center.

The center of the hyperbola is at the point (h,k), which in this case is (1,9). The vertices are located a distance a from the center along the x-axis, where a² = 49, so a = 7. Thus, the vertices are at (1±7,9) or (-6,9) and (8,9).

To find the foci, we need to calculate the focal distance using the equation c² = a² + b², where a² = 49 and b² = 25. Therefore, c = √(49+25), which simplifies to c = √74. The foci are then at (1±√74,9).

Lastly, the asymptotes of a hyperbola are straight lines that pass through the center and have slopes equal to ±b/a. Here, b/a = ±(5/7), thus the equations of the asymptotes are y-9 = ±(5/7)(x-1).

User Moataz Elmasry
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