Question

question 5 of 10 which of the following is the graph of this hyperbola?
\\(\frac{(x + 1)^2}{16}-\frac{(y - 2)^2}{144}=1\\)

Explanation:

Step1: Identify the standard - form of hyperbola

The standard form of a hyperbola with a horizontal transverse axis is $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, and with a vertical transverse axis is $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$. The given equation is $\frac{(x + 1)^2}{16}-\frac{(y - 2)^2}{144}=1$, so it has a horizontal transverse axis with center $(h,k)=(-1,2)$, $a^2 = 16$ (so $a = 4$) and $b^2=144$ (so $b = 12$).

Step2: Analyze the vertices

For a hyperbola of the form $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, the vertices are at $(h\pm a,k)$. Substituting $h=-1$, $a = 4$ and $k = 2$, we get the vertices at $(-1+4,2)=(3,2)$ and $(-1 - 4,2)=(-5,2)$.

Step3: Match with the graphs

We look for a hyperbola with a horizontal transverse axis, center at $(-1,2)$ and vertices at $(-5,2)$ and $(3,2)$.

Answer:

C.