Question

ch. then sketch the graph.

1. (\frac{y^2}{4} - \frac{x^2}{4} = 1)

Explanation:

Step1: Identify the hyperbola type

The equation is \(\frac{y^{2}}{4}-\frac{x^{2}}{4} = 1\), which is a vertical hyperbola in the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\). Here, \(a^{2}=4\) so \(a = 2\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the center

For the hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), the center is at \((h,k)=(0,0)\) (since there are no shifts in \(x\) or \(y\) from the standard form).

Step3: Find the vertices

For a vertical hyperbola, vertices are at \((h,k\pm a)\). Substituting \(h = 0\), \(k = 0\), and \(a = 2\), the vertices are \((0,0 + 2)=(0,2)\) and \((0,0-2)=(0,-2)\).

Step4: Find the foci

First, calculate \(c\) using the relationship \(c^{2}=a^{2}+b^{2}\). So \(c^{2}=4 + 4=8\), thus \(c=\sqrt{8}=2\sqrt{2}\). Foci for vertical hyperbola are at \((h,k\pm c)\), so \((0,0\pm2\sqrt{2})=(0,2\sqrt{2})\) and \((0,-2\sqrt{2})\).

Step5: Find the asymptotes

For vertical hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), asymptotes are \(y=\pm\frac{a}{b}x\). Since \(a = b = 2\), \(\frac{a}{b}=1\), so asymptotes are \(y=\pm x\).

Step6: Sketch the graph

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,2)\) and \((0,-2)\).
3. Draw the asymptotes \(y = x\) and \(y=-x\) (lines through the center with slopes \(1\) and \(- 1\)).
4. Sketch the hyperbola opening up and down, approaching the asymptotes, with vertices at \((0,\pm2)\).

Answer:

• Center: \((0,0)\)
• Vertices: \((0,2)\), \((0, - 2)\)
• Foci: \((0,2\sqrt{2})\), \((0,-2\sqrt{2})\)
• Asymptotes: \(y = x\), \(y=-x\)
• Graph: Vertical hyperbola centered at origin, vertices on \(y\)-axis at \((0,\pm2)\), asymptotes \(y=\pm x\), opening up and down.