please show work so i can understand how you solved it. thanks!

## write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4.

# 1 Answer

Given: center = (h, k) = (-2, -4)

focus = (-2, 6)

eccentricity = c/a = (√(a^{2} + b^{2}))/a = 5/4

First notice that both the center and the focus have the same x coordinate (x= -2), which means that the transverse axis is parallel to the y-axis (i.e., the transverse axis is vertical). From this we know that the standard form of this hyperbola is as follows:

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

Second, since the eccentricity is equal to c/a, then a=4 and c = 5. From this we can find b using the fact that c equal the square root of the sum of a^{2} and b^{2}:

c = √(a^{2} + b^{2})

5 = √(4^{2} + b^{2})

5 = √(16 + b^{2})

Square both sides of this equation then subtract 16 from both sides:

25 = 16 + b^{2}

25 - 16 = b^{2}

9 = b^{2}

3 = b

Using the values that we've found for a and b, as well as the center ((h, k) = (-2, -4)) given and the standard form of a hyperbola with a vertical transverse axis, we can generate the equation of the hyperbola in standard form:

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

(y - (-4))^{2}/3^{2} - (x - (-2))^{2}/4^{2} = 1

(y + 4)^{2}/9 - (x + 2)^{2}/16 = 1

## Comments

Hi Tamara. Equation of the Hyperbola with Center (h,k) and Focal Axis Parallel to y-axis has form:

(y-k)

^{2}/a^{2}- (x-h)^{2}/b^{2}= 1Comment