Questions: Nr → xt θ ○ x

Transcript text: $\mathrm{Nr} \rightarrow \mathrm{xt}$ $\theta \bigcirc x$

Solution

Solution Steps

Step 1: Rewrite the equation in standard form

The standard form of a hyperbola centered at the origin is either $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. The given equation is $\frac{y^2}{16} - \frac{x^2}{4} = 1$. This is already in standard form, with $a^2 = 16$ and $b^2 = 4$, so $a=4$ and $b=2$. Since the $y^2$ term is positive, the hyperbola opens vertically.

Step 2: Identify the center, vertices, and asymptotes

The center of the hyperbola is at $(0,0)$. The vertices are at $(0, \pm a)$, so they are located at $(0,4)$ and $(0, -4)$. The asymptotes are given by $y = \pm \frac{a}{b}x$, which in this case are $y = \pm \frac{4}{2}x$, simplifying to $y= \pm 2x$.

Step 3: Sketch the graph

Plot the center $(0,0)$ and vertices $(0,4)$ and $(0,-4)$. Draw the asymptotes $y = 2x$ and $y = -2x$. Sketch the hyperbola branches opening vertically, passing through the vertices and approaching the asymptotes.

Final Answer:

The hyperbola is centered at $(0,0)$, has vertices at $(0,4)$ and $(0,-4)$, and asymptotes $y=2x$ and $y=-2x$. It opens vertically. A sketch of the hyperbola would show curves passing through the vertices and approaching the asymptotes.

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Questions: Nr → xt θ ○ x

Solution

Solution Steps

Final Answer:

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