Question

question 8 of 10 the foci for the hyperbola $\frac{(y - 2)^2}{16}-\frac{(x + 1)^2}{144}=1$ are $(-1,2 + 4sqrt{10})$ and $(-1,2 - 4sqrt{10})$. a. true b. false

Explanation:

Step1: Identify the standard - form of hyperbola

The given hyperbola equation is $\frac{(y - 2)^2}{16}-\frac{(x + 1)^2}{144}=1$, which is of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, where $(h,k)=(-1,2)$, $a^2 = 16$ (so $a = 4$) and $b^2=144$ (so $b = 12$).

Step2: Calculate $c$

For a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, the relationship between $a$, $b$, and $c$ is $c^2=a^2 + b^2$. Substitute $a = 4$ and $b = 12$ into the formula: $c^2=16 + 144=160$, so $c=\sqrt{160}=4\sqrt{10}$.

Step3: Find the foci

The foci of a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ are given by the points $(h,k\pm c)$. Here, $h=-1$, $k = 2$ and $c = 4\sqrt{10}$, so the foci are $(-1,2 + 4\sqrt{10})$ and $(-1,2-4\sqrt{10})$.

Answer:

A. True