Question

write an equation for the hyperbola shown in the graph.
the equation for the hyperbola above is □ = 1.
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Determine the standard form

The standard form of a hyperbola with a vertical transverse axis is \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\). The center of the hyperbola is \((0,0)\).

Step2: Find the value of \(a\)

From the graph, the distance from the center \((0,0)\) to a vertex is \(a\). Let's assume a vertex point (from the graph's construction with the dashed - line rectangle). If we consider the vertical side of the rectangle related to the vertices, and using the fact that for the hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), the asymptotes are given by \(y=\pm\frac{a}{b}x\).
The asymptotes are \(y = 2x\) and \(y=-2x\), so \(\frac{a}{b}=2\), i.e., \(a = 2b\).

Step3: Use the relationship from the standard form

The equation of the hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) can be rewritten in terms of \(b\) since \(a = 2b\). Substituting \(a = 2b\) into the equation, we get \(\frac{y^{2}}{4b^{2}}-\frac{x^{2}}{b^{2}}=1\).
Let's use the fact that if we consider the rectangle (a fundamental rectangle for the hyperbola) which has sides \(2a\) (vertical) and \(2b\) (horizontal). From the graph (by the structure of the asymptotes and the hyperbola - rectangle relationship), we can also note that for the hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), another way:
The general form of the asymptotes for \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\) is \(y=\pm\frac{a}{b}x\). Given \(y = 2x\) and \(y=-2x\), so \(\frac{a}{b}=2\). Let's assume a point on the hyperbola (using the rectangle - hyperbola relation). The standard form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\).
If we take \(x = b\), then \(y=a\). Since \(\frac{a}{b}=2\), and using the fact that for the hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), we can rewrite it as \(\frac{y^{2}}{4}-\frac{x^{2}}{1}=1\) (because if we let \(b = 1\), then \(a=2\)).

Answer:

\(\frac{y^{2}}{4}-x^{2}\)