Question

a) determine the equations of the asymptotes for the equation
b) graph the equation
\\(\frac{x^2}{25} - \frac{y^2}{1} = 1\\)

a) choose the correct answer below.
\\(\bigcirc\\) a. \\(y = 25x, y = -25x\\)
\\(\bigcirc\\) b. \\(y = \frac{1}{5}x, y = -\frac{1}{5}x\\)
\\(\bigcirc\\) c. \\(y = \frac{1}{25}x, y = -\frac{1}{25}x\\)
\\(\bigcirc\\) d. \\(y = 5x, y = -5x\\)

b) choose the correct graph below.
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.
\\(\bigcirc\\) c.
\\(\bigcirc\\) d.

Explanation:

Part (a)

Step1: Recall Hyperbola Asymptote Formula

For a hyperbola of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y=\pm\frac{b}{a}x\).

Step2: Identify \(a\) and \(b\) from the Given Equation

The given equation is \(\frac{x^2}{25}-\frac{y^2}{1}=1\). Comparing with \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), we have \(a^2 = 25\) (so \(a = 5\)) and \(b^2=1\) (so \(b = 1\)).

Step3: Calculate the Asymptotes

Substitute \(a = 5\) and \(b = 1\) into the asymptote formula \(y=\pm\frac{b}{a}x\). We get \(y=\pm\frac{1}{5}x\), which is \(y=\frac{1}{5}x\) and \(y =-\frac{1}{5}x\).

Step1: Analyze the Hyperbola Equation

The equation \(\frac{x^2}{25}-\frac{y^2}{1}=1\) is a horizontal hyperbola (since the \(x^2\) term is positive) with \(a = 5\) (so the vertices are at \((\pm5,0)\)) and asymptotes \(y=\pm\frac{1}{5}x\).

Step2: Match with the Graphs

• A horizontal hyperbola opens left and right, with vertices on the \(x\)-axis at \((\pm5,0)\) and asymptotes with slope \(\pm\frac{1}{5}\).
• Looking at the options, the graph that has a horizontal hyperbola (opening left - right) with vertices near \((\pm5,0)\) and the correct asymptote slope is the one that matches the equation. From the visual options (even though we can't see the exact graphs, based on the equation), the correct graph should be the one with a horizontal hyperbola centered at the origin, vertices at \((\pm5,0)\), and asymptotes \(y=\pm\frac{1}{5}x\). Assuming the options are labeled, and from the context, the correct graph (let's assume the first graph in the options for horizontal hyperbola with \(a = 5\)) - but since we need to pick based on the standard form, the graph with horizontal hyperbola (opening left and right) with vertices at \(\pm5\) on the \(x\)-axis. (If we consider the options, the graph that is a horizontal hyperbola with vertices at \((\pm5,0)\) and the correct asymptotes. Let's assume the correct graph is the one labeled as, say, the first one (A) if it has the correct features, but since the original problem's graphs: the hyperbola \(\frac{x^2}{25}-\frac{y^2}{1}=1\) has vertices at \((\pm5,0)\) and opens left - right. So the correct graph should be the one with a horizontal hyperbola (opening left and right) with vertices at \((\pm5,0)\).

Answer:

B. \(y=\frac{1}{5}x,y =-\frac{1}{5}x\)

Part (b)