91Ó°ÊÓ

To understand hyperbolas, it's important to start with the equation in standard form. The general equation for a hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This equation gives us a detailed description of the hyperbola's shape and position in the coordinate plane.
- **\(a^2\)** and **\(b^2\)** are constants that define how stretched the hyperbola is along the x-axis and y-axis respectively.- The signs tell us how the hyperbola opens: '-' means the hyperbola opens along the x-direction, while a '+' after would open along the y-direction.
Transforming equations into this standard form helps easily identify properties of the hyperbola, such as its center and vertices.

Completing the square is a crucial step in rewriting equations. It allows us to express quadratic expressions in a more manageable form.
- Begin by grouping x-terms and y-terms separately.- For the x-terms in the equation \(5x^2 + 10x\): 1. Factor out the coefficient of \(x^2\), which is 5. 2. Rewrite as: \(5(x^2 + 2x)\). 3. Completing the square inside gives \(5((x + 1)^2 - 1)\), leading to \(5(x + 1)^2 - 5\).- Similarly, for the y-terms \(-7y^2 + 28y\), factor out -7 and complete the square to get \(-7(y - 2)^2 + 28\).
Incorporate these steps to simplify and transform the equation to the standard form, making it easier to identify the center and vertices.

The center of a hyperbola is a key point that helps define its position. Once we transform the hyperbola into its standard form, identifying the center becomes straightforward.
- For an equation in the form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), the center is \((h, k)\).- In our example, transforming the equation provides \((x + 1)^2\) and \((y - 2)^2\), leading to the center \((-1, 2)\).
The center acts as a reference point for describing other features of the hyperbola, such as its vertices and asymptotes.

Vertices are essential points on a hyperbola that indicate its extent along its transverse axis. From the standard form equation, calculating these becomes easier.
- The vertices are located at a distance \(a\) from the center along the axis of symmetry of the hyperbola.- In our example, with \(a^2 = 7\), we identify that \(a = \sqrt{7}\).- Thus, the vertices can be found at \((-1 \pm \sqrt{7}, 2)\), reflecting their positions along the x-axis given our equation.
Understanding the position of the vertices is essential in sketching the hyperbola and assessing its basic shape.