Question

find all vertices and foci of the hyperbola given by

\frac{(x + 2)^2}{25} - \frac{(y - 4)^2}{39} = 1

enter your answers as ordered pairs separated by a comma.

Explanation:

Step1: Identify the standard form

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\). For the given hyperbola \(\frac{(x + 2)^2}{25}-\frac{(y - 4)^2}{39}=1\), we have \(h=-2\), \(k = 4\), \(a^2=25\) (so \(a = 5\)), \(b^2=39\).

Step2: Find \(c\)

Use the formula \(c^2=a^2 + b^2\). Substitute \(a^2 = 25\) and \(b^2=39\), then \(c^2=25 + 39=64\), so \(c = 8\).

Step3: Find vertices

For a hyperbola of the form \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), the vertices are \((h\pm a,k)\). Substitute \(h=-2\), \(a = 5\), \(k = 4\). The vertices are \((-2+5,4)=(3,4)\) and \((-2 - 5,4)=(-7,4)\).

Step4: Find foci

For a hyperbola of the form \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\), the foci are \((h\pm c,k)\). Substitute \(h=-2\), \(c = 8\), \(k = 4\). The foci are \((-2+8,4)=(6,4)\) and \((-2-8,4)=(-10,4)\).

Answer:

Vertices: \((3,4),(-7,4)\); Foci: \((6,4),(-10,4)\)