Question

question 20 of 20
what are the coordinates of the vertices of the conic section shown below?
\\(\frac{(x + 2)^2}{16}-\frac{(y - 3)^2}{9}=1\\)
a. (-2,-1) and (-2,7)
b. (-2,3) and (6,3)
c. (-6,3) and (2,3)
d. (-2,-7) and (-2,1)

Explanation:

Step1: Identify the form of hyperbola

The given equation $\frac{(x + 2)^2}{16}-\frac{(y - 3)^2}{9}=1$ is of the form $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, which represents a horizontal - hyperbola with center $(h,k)$. Here $h=-2,k = 3,a^2 = 16$ (so $a = 4$), and $b^2=9$ (so $b = 3$).

Step2: Find the vertices

For a horizontal hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, the vertices are given by the points $(h - a,k)$ and $(h + a,k)$. Substitute $h=-2,k = 3,a = 4$ into the formula.
For the first vertex: $x=h - a=-2-4=-6,y = k = 3$.
For the second vertex: $x=h + a=-2 + 4=2,y = k = 3$.

Answer:

C. (-6,3) and (2,3)