Question

for \\( \frac{x^{2}}{36}-\frac{y^{2}}{49}=1 \\),
(a) identify the center.
(b) identify the vertices.
(c) identify the foci.
(d) write equations for the asymptotes.
(e) graph the hyperbola.
give exact answers. express numbers in simplest form.

Explanation:

Step1: Find the center

The standard form of a hyperbola is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\). For the given equation \(\frac{x^{2}}{36}-\frac{y^{2}}{49}=1\), we have \(h = 0,k = 0\).

Step2: Find the vertices

Since \(a^{2}=36\), then \(a = 6\). The vertices of a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \((\pm a,0)\). So the vertices are \((6,0)\) and \((-6,0)\).

Step3: Find the foci

We know that \(c^{2}=a^{2}+b^{2}\). Given \(a^{2}=36\) and \(b^{2}=49\), then \(c^{2}=36 + 49=85\), so \(c=\sqrt{85}\). The foci are \((\pm\sqrt{85},0)\).

Step4: Find the asymptotes

The equations of the asymptotes for a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \(y=\pm\frac{b}{a}x\). Since \(a = 6\) and \(b = 7\), the asymptotes are \(y=\pm\frac{7}{6}x\).

Step5: Graph the hyperbola

1. Plot the center at the origin \((0,0)\).
2. Plot the vertices at \((6,0)\) and \((-6,0)\).
3. Sketch the asymptotes \(y=\frac{7}{6}x\) and \(y =-\frac{7}{6}x\) as dashed lines.
4. Draw the hyperbola approaching the asymptotes.

Answer:

(a) Center: \((0,0)\)
(b) Vertices: \((6,0)\) and \((-6,0)\)
(c) Foci: \((\sqrt{85},0)\) and \((-\sqrt{85},0)\)
(d) Asymptotes: \(y=\frac{7}{6}x\) and \(y =-\frac{7}{6}x\)
(e) Graph: (Describe as per the steps above with center at origin, vertices on x - axis, asymptotes \(y=\pm\frac{7}{6}x\) and hyperbola approaching them)