Hyperbola Vertices Coordinates
Find the coordinates of either of the vertices of the hyperbola \[16x^2+16x-4y^2-20y-85=0.\](Enter your answer as an ordered pair. Enter the coordinates of one of the vertices, not both.)
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
To find the standard form for the equation of the hyperbola, we complete the square in both variables: \[\begin{aligned} 16(x^2+x) - 4(y^2+5y) - 85 &= 0 \\ 16(x^2+x+\tfrac14)-4(y^2+5y+\tfrac{25}4) - 85 &= 4 - 25 \\ 16(x+\tfrac12)^2 - 4(y+\tfrac52)^2 &= 64 \\ \frac{(x+\tfrac12)^2}{4} - \frac{(y+\tfrac52)^2}{16} &= 1.\end{aligned}\]Therefore, the center of the hyperbola is the point $ \left(-\tfrac12, -\tfrac52\right) $. The vertices lie to the left and right of the center, and the distance from the center to each vertex is $ \sqrt{4} = 2 $. Thus, the vertices have coordinates \[\left(-\tfrac12 \pm 2,-\tfrac52\right) = \boxed{\left(\tfrac32, -\tfrac52\right)} \text{ and } \boxed{\left(-\tfrac52, -\tfrac52\right)}.\](Either point is a correct answer to this problem.)
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