Problem 19 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (0,4),(0,0) passes through the point \((\sqrt{5},-1)\)

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola with given vertices and passing through the point is \( \frac{(y-2)^2}{16} - \frac{2x^2}{5} = 1 \).

Step 1: Calculating Semi-axis Lengths

First, calculate the distance between the given vertices (0,4) and (0,0) to find the length of the semi-major axis. This is done by using the distance formula, which in this case simplifies to subtracting the y-values. Therefore, the length of the semi-major axis is \( a = |4-0| = 4 \). The center of the hyperbola is the midpoint of the vertices, which is \( (0, 2) \).

Step 2: Deriving the Auxiliary Equation from the Given Point

Now using the standard form equation of a vertical hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) and the known values of center \( (h, k) = (0, 2) \) and the semi-major axis \( a = 4 \), we substitute using point \((\sqrt{5}, -1)\) and solve for \( b^2 \). We get \( \frac{(-1-2)^2}{4^2} - \frac{\sqrt{5}^2}{b^2} = 1 \). Solving this yields \( b^2 = \frac{5}{2} \).

Step 3: Forming the Equation of the Hyperbola

Using the final values of \( a \), \( b \) and center \( h, k \), we can form the equation of the hyperbola. We substitute these into the original equation to obtain: \( \frac{(y-2)^2}{4^2} - \frac{x^2}{\frac{5}{2}} = 1 \). Simplifying this, we get the standard form of the equation to be \( \frac{(y-2)^2}{16} - \frac{2x^2}{5} = 1 \).

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are significant as they help to define the shape and orientation of the hyperbola on a coordinate plane. As stated in the original problem, we have two vertices: (0,4) and (0,0). These vertices indicate that the hyperbola is oriented vertically because they share the same x-coordinate. Linear segments joining the vertices form part of the hyperbola's transverse axis. The midpoint of this line segment is the center of the hyperbola, which is calculated by taking the average of the y-coordinates since the x-values are the same, therefore yielding a center at (0,2). Together, the vertices and center help us to ascertain the orientation and foundational shape of the hyperbola.

Distance Formula

The distance formula is a crucial tool in finding the length between two points in a coordinate plane, which is often applied in various problems involving geometrical shapes. To find the distance between the vertices of the hyperbola given in this exercise, we use the distance formula which is: \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
In our given scenario, since the x-coordinates are identical (both being zero), this simplifies to subtracting the y-coordinates: \( |4-0| \) which gives us a distance of 4.
This distance between vertices basically represents the length of the semi-major axis of the hyperbola thanks to the positioning of the vertices. Thus, it is crucial to determining other properties of the hyperbola.

Standard Form of a Hyperbola

The standard form of a hyperbola is an essential formulation that allows us to express its equation neatly and orderly. For a vertically oriented hyperbola, as seen in this particular exercise where the vertices (0,4) and (0,0) lie on the same vertical line, the standard form equation is given by: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]where \((h, k)\) is the center of the hyperbola, and \(a\) and \(b\) are the semi-major and semi-minor axes respectively. In the solved example, the center \((h, k)\) is \((0, 2)\), \(a = 4\) (previously calculated), and \(b^2\) was derived as \(\frac{5}{2}\) by substituting the point \((\sqrt{5}, -1)\) into the equation.
Once these values are substituted back into the standard form, simplifying yields \( \frac{(y-2)^2}{16} - \frac{2x^2}{5} = 1 \), providing a concise representation of the hyperbola's equation in standard form.

Semi-major Axis

The semi-major axis is a fundamental element of any hyperbola, representing half of the length of its transverse axis. In hyperbolas, the transverse axis is the line that connects the two vertices. In this problem, we found the semi-major axis by measuring the distance between vertices (via the distance formula). Since we obtained this distance as 4, the semi-major axis, denoted as \(a\), is directly derived to be of length 4.
Understanding the semi-major axis is crucial, as it sets the scale for the hyperbola's shape. Moreover, \(a\) is used in the calculation of other significant attributes of the hyperbola, like eccentricity and focal distance, and crucially forms a part of the equation of the hyperbola, in this case with \(a^2 = 16\). Thus, the semi-major axis is not just a measurement but a determinant of the hyperbola's geometry.