Question

find all vertices and foci of the hyperbola given by
\\( \frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 28 } = 1 \\)
enter your answers as ordered pairs separated by a comma.
vertices

Explanation:

Step1: Identify the form of hyperbola

The standard form of a hyperbola is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\) (opens left - right). For the given hyperbola \(\frac{x^{2}}{36}-\frac{y^{2}}{28}=1\), we have \(a^{2}=36\), so \(a = 6\), and \(b^{2}=28\).

Step2: Find the vertices

The vertices of a hyperbola of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \((\pm a,0)\). Substituting \(a = 6\), the vertices are \((6,0)\) and \((- 6,0)\).

Step3: Calculate \(c\)

We use the relationship \(c^{2}=a^{2}+b^{2}\). Substituting \(a^{2}=36\) and \(b^{2}=28\), we get \(c^{2}=36 + 28=64\), so \(c = 8\).

Step4: Find the foci

The foci of a hyperbola of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \((\pm c,0)\). Substituting \(c = 8\), the foci are \((8,0)\) and \((-8,0)\).

Answer:

Vertices: \((6,0),(-6,0)\); Foci: \((8,0),(-8,0)\)