Question

Sketch the hyperbola defined by the given equation. Label the center and foci. \((y-4)^{2}-\frac{(x+1)^{2}}{25}=1\)

Short answer

Center: (-1, 4); Foci: (-1, 4±√26); Asymptotes: y = 4±1/5(x+1).

Step-by-step solution

Identify the Form of the Equation

The given equation is \((y-4)^{2} - \frac{(x+1)^{2}}{25} = 1\). This equation is in the form of \((y-k)^2 - \frac{(x-h)^2}{a^2} = 1\), which represents a vertical hyperbola. The center of the hyperbola is \((h, k)\), so here, the center is \((-1, 4)\).

Determine the Values of 'a' and 'b'

In a hyperbola of the form \((y-k)^2 - \frac{(x-h)^2}{a^2} = 1\), \(a^2\) is the denominator of the \(x\) term. Here, \(a^2 = 25\) gives \(a = 5\). For a vertical hyperbola, we compare \(b^2\) to 1 since the \((y-k)^2\) term is on its own. Thus, \(b^2 = 1\) gives \(b = 1\).

Calculate the Distance to the Foci

The distance to the foci from the center of a vertical hyperbola is given by \(c = \sqrt{a^2 + b^2}\). Substituting \(a^2 = 25\) and \(b^2 = 1\), we get \(c = \sqrt{25 + 1} = \sqrt{26}\). Hence, the foci are \((-1, 4 + \sqrt{26})\) and \((-1, 4 - \sqrt{26})\).

Sketch the Asymptotes and Hyperbola

For a vertical hyperbola, the equations of the asymptotes are \(y = k \pm \frac{b}{a}(x - h)\). Substituting \(b = 1\), \(a = 5\), \(k = 4\), and \(h = -1\), the asymptote equations are \(y = 4 \pm \frac{1}{5}(x + 1)\). Draw the center at \((-1, 4)\), sketch the asymptotes, and plot the hyperbola opening upwards and downward around these asymptotes.

Key concepts

Center of Hyperbola

The center of a hyperbola is essentially its midpoint. It is a crucial part of defining the hyperbola's position on a graph. For the equation \[(y-4)^2 - \frac{(x+1)^2}{25} = 1\]we identify the center by comparing it to the general form of a vertical hyperbola: \[(y-k)^2 - \frac{(x-h)^2}{a^2} = 1\]Here, \(h\) and \(k\) are the coordinates of the center. By looking at the given equation, we find:

• \(h = -1\)
• \(k = 4\)

Thus, the center of this hyperbola is \((-1, 4)\). This point is the origin of symmetry for our hyperbola, meaning that it will mirror exactly at this point.

Foci of Hyperbola

The foci (plural of focus) are critical features of a hyperbola, lying on the axis through the center, which involves where each "arm" of the hyperbola appears to aim toward. The distance to each focus from the center is given by the formula:\[c = \sqrt{a^2 + b^2}\]For this specific hyperbola:

• \(a^2 = 25\), so \(a = 5\)
• \(b^2 = 1\), thus \(b = 1\)

With these values, we calculate:\[c = \sqrt{25 + 1} = \sqrt{26}\]The hyperbola's foci are then positioned at:

• \((-1, 4 + \sqrt{26})\)
• \((-1, 4 - \sqrt{26})\)

These points lie along the line parallel to the \(y\)-axis, passing through the hyperbola's center.

Equation of Hyperbola

Understanding the equation of a hyperbola is key in graphing it precisely. For a vertical hyperbola, the standard form is:\[(y-k)^2 - \frac{(x-h)^2}{a^2} = 1\]Our given equation matches this template: \[(y-4)^2 - \frac{(x+1)^2}{25} = 1\]Here’s how each part influences the shape and position:

• The \((y-k)^2\) term signifies it's a vertical hyperbola, as the \(y\) is separated on one side.
• The \(h\) and \(k\) values \((-1, 4)\) set our center point.
• \(a^2 = 25\) provides the \(x\)-intercepts and influences how "stretched" it is along x.
• The negative sign reflects how the \((x+1)^2\) term modifies the vertical setting.

Understanding this structure allows you to identify key features, like orientation and symmetry.

Sketching Asymptotes

Asymptotes are imaginary lines that the hyperbola approaches but never actually touches. They provide a frame that guides the hyperbola's overall shape. For vertical hyperbolas, the asymptotes' equations look like:\[y = k \pm \frac{b}{a}(x-h)\]With our specific equation, we use:

• \(k = 4\)
• \(h = -1\)
• \(a = 5\)
• \(b = 1\)

Substituting these into our asymptote formula, we find:\[y = 4 \pm \frac{1}{5}(x + 1)\]It's useful to sketch these asymptotes on a graph:

• The slope \(\frac{1}{5}\) tells us precisely how "tilted" they are.
• Draw these lines in a way that they intersect at the center \((-1, 4)\).

These lines paint the boundaries within which our hyperbola will snugly fit, opening in the traditionally vertical manner.