Question

graph the hyperbola given by the following equation\\(\frac{(y + 2)^2}{25} - \frac{x^2}{11} = 1\\)

Explanation:

Step1: Identify Hyperbola Type

The equation is \(\frac{(y + 2)^2}{25}-\frac{x^2}{11}=1\), which matches the vertical transverse axis hyperbola standard form \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\). Here, \(h = 0\), \(k=-2\), \(a^2 = 25\) (so \(a = 5\)), \(b^2=11\) (so \(b=\sqrt{11}\approx3.316\)).

Step2: Find Center

The center \((h,k)\) is \((0, -2)\) (from \(h = 0\), \(k=-2\) in the standard form).

Step3: Find Vertices

For a vertical hyperbola, vertices are at \((h,k\pm a)\). So vertices are \((0, -2 + 5)=(0, 3)\) and \((0, -2 - 5)=(0, -7)\).

Step4: Find Asymptotes

The equations of asymptotes for vertical hyperbola are \(y - k=\pm\frac{a}{b}(x - h)\). Substituting values: \(y + 2=\pm\frac{5}{\sqrt{11}}x\) (or rationalized as \(y + 2=\pm\frac{5\sqrt{11}}{11}x\)).

Step5: Sketch the Hyperbola

1. Plot the center \((0, -2)\).
2. Plot the vertices \((0, 3)\) and \((0, -7)\).
3. Draw the asymptotes (dashed lines) using the asymptote equations.
4. Sketch the two branches of the hyperbola, opening up and down, approaching the asymptotes, and passing through the vertices.

Answer:

To graph \(\boldsymbol{\frac{(y + 2)^2}{25}-\frac{x^2}{11}=1}\):

• Center: \((0, -2)\)
• Vertices: \((0, 3)\) and \((0, -7)\)
• Asymptotes: \(y + 2=\pm\frac{5\sqrt{11}}{11}x\)
• Sketch branches opening vertically (up/down), passing through vertices, approaching asymptotes.

(For a visual graph, plot these key points/lines and draw the hyperbola accordingly.)