Question

graph the hyperbola given by the following equation
\\( \frac { ( y - 1 ) ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 16 } = 1 \\)

Explanation:

Step1: Identify the standard form

The standard form of a hyperbola is $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ for a hyperbola that opens up and down.

Step2: Determine center, \(a\), \(b\)

For the equation $\frac{(y - 1)^2}{9}-\frac{x^2}{16}=1$, the center \((h,k)=(0,1)\). Since \(a^2 = 9\), then \(a = 3\). Since \(b^2=16\), then \(b = 4\).

Step3: Find vertices

The vertices of a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ are \((h,k\pm a)\). So the vertices are \((0,1 + 3)=(0,4)\) and \((0,1-3)=(0,-2)\).

Step4: Find asymptotes

The equations of the asymptotes for a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ are \(y - k=\pm\frac{a}{b}(x - h)\). Substituting \(h = 0,k = 1,a = 3,b = 4\), we get \(y-1=\pm\frac{3}{4}x\) or \(y=\frac{3}{4}x + 1\) and \(y=-\frac{3}{4}x+1\).

To graph:

• Plot the center at \((0,1)\).
• Plot the vertices at \((0,4)\) and \((0,-2)\).
• Sketch the asymptotes \(y=\frac{3}{4}x + 1\) and \(y=-\frac{3}{4}x + 1\).
• Draw the hyperbola approaching the asymptotes.

Answer:

The hyperbola has center \((0,1)\), vertices \((0,4)\) and \((0,-2)\), and asymptotes \(y=\frac{3}{4}x + 1\) and \(y=-\frac{3}{4}x + 1\). Graph by plotting these key elements and drawing the hyperbola branches approaching the asymptotes.