Problem 41 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (±1,0)\(;\) asymptotes: \(y=\pm 5 x\)

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \(x^2 - y^2/25 = 1\).

Step 1: Find The Value of 'a'

The value of \(a\), the distance from the center of the hyperbola to each vertex, is given by the x coordinate of the vertex. In this case, as the vertices are at (±1,0), the value of \(a\) is 1.

Step 2: Find the Value of 'b'

The value of \(b\), the distance from the center to each asymptote, is found by observing the slope of the asymptotes. The asymptotes are given as \(y = \pm 5x\), therefore the slope is \(\pm 5\). This slope is also equal to the ratio \(b/a\). Since \(a\) is 1, \(b\) is also 5.

Step 3: Formulate the Equation

Now that we have the values of \(a\) and \(b\), we can formulate the equation of the hyperbola. Since the vertices are on the \(x\)-axis, we know this is a horizontal hyperbola. Hence its standard form equation will be \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Here, h and k represent the coordinates of the center of the hyperbola; as it is not given explicitly, we can assume it to be at origin i.e., (0, 0). So the equation becomes \(\frac{x^2}{1^2} - \frac{y^2}{5^2} = 1\), which simplifies to \(x^2 - y^2/25 = 1\).