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Write the equation of the hyperbola centered at (-7,8). The length of the vertical transverse axis is 2 and the length of the conjugates axis is 11

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Final answer:

The equation of the hyperbola centered at (-7,8) with a 2-unit vertical transverse axis and an 11-unit conjugate axis is (y - 8)^2 - (x + 7)^2 / 30.25 = 1.

Step-by-step explanation:

To write the equation of a hyperbola centered at (-7,8) with the length of the vertical transverse axis being 2 and the length of the conjugate axis being 11, we first need to understand the standard form of a hyperbola's equation.

For a hyperbola with a vertical transverse axis, the equation is of the form:

(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1

Where (h, k) is the center of the hyperbola, a is half the length of the transverse axis (the distance from the center to a vertex), and b is half the length of the conjugate axis.

Given that the transverse axis is 2 units long, a would be 1 (half of 2). With an 11-unit-long conjugate axis, b would be 5.5 (half of 11). The center of the hyperbola is given as (-7, 8). Plugging these values into the standard equation gives:

(y - 8)^2 / 1^2 - (x + 7)^2 / 5.5^2 = 1

Which simplifies to:

(y - 8)^2 - (x + 7)^2 / 30.25 = 1

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