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question 7 of 10 the vertices for the hyperbola $\frac{(x - 1)^2}{25}-\…

Question

question 7 of 10
the vertices for the hyperbola $\frac{(x - 1)^2}{25}-\frac{(y + 3)^2}{9}=1$ are $(1,0)$ and $(1,-6)$.
a. true
b. false

Explanation:

Step1: Identify the form of hyperbola

The given hyperbola equation $\frac{(x - 1)^2}{25}-\frac{(y + 3)^2}{9}=1$ is of the form $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, where $(h,k)$ is the center, $a^2 = 25$ so $a = 5$, and the center is $(h,k)=(1,-3)$.

Step2: Find the vertices

For a hyperbola of the form $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, the vertices are given by $(h\pm a,k)$. Here, $h = 1,k=-3,a = 5$. The vertices are $(1 + 5,-3)=(6,-3)$ and $(1-5,-3)=(-4,-3)$.

Answer:

B. False