91Ó°ÊÓ

A hyperbola is a type of conic section that forms when a plane intersects both nappes (the upper and lower parts) of a double cone. The resulting shape consists of two separate curves. A hyperbola has a distinct feature where the difference in distances from any point on the hyperbola to two fixed points called foci is constant. Hyperbolas are represented by equations of the form: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] or \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]. Here are some key characteristics:

• Center: The midpoint of the segment connecting the two foci.
• Vertices: Points where the hyperbola intersects its transverse axis.
• Asymptotes: Lines that the hyperbola approaches but never touches.

In our exercise, the equation \[ -\frac{(x-2)^2}{100} + \frac{y^2}{1} = 1 \] represents a hyperbola since it fits the standard form and includes a negative sign indicating the curve opens left and right.

Completing the square is an algebraic method used to transform a quadratic equation into a perfect square trinomial, making it easier to solve or rewrite. This method is crucial for converting quadratic equations to standard conic forms.

To complete the square for a term like \(ax^2 + bx\):

• First, factor out the coefficient of \(x^2\) if it isn't 1.
• Divide the coefficient of \(x\) by 2, then square it.
• Add and subtract this squared term inside the parenthesis.

Here’s how we apply it to \(x^2 - 4x\):

1. Factor out any common coefficients (in this case, it's already simplified).
2. Take half the coefficient of \(x\) (-4), which is -2, and square it to get 4.
3. Rewrite \(x^2 - 4x + 4 - 4\), simplifying to \((x-2)^2 - 4\).

This technique is used in our step-by-step solution to simplify and rewrite the equation to identify the conic section.

The standard form of a conic section equation allows for easier identification and analysis. Each conic type (ellipse, hyperbola, parabola, and circle) has a distinct standard form:

• Circle: \((x-h)^2 + (y-k)^2 = r^2\)
• Ellipse: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• Parabola: \((x-h)^2 = 4p(y-k)\) or \((y-k)^2 = 4p(x-h)\)
• Hyperbola: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)

Our goal is to convert any given conic equation to its respective standard form. This involves manipulating and simplifying terms, often through methods like completing the square. In the provided exercise, starting with \(100y^2 + 4x = x^2 + 104\), we used completing the square on the \(x\) terms to eventually rewrite it in the hyperbola's standard form: \[-\frac{(x-2)^2}{100} + \frac{y^2}{1} = 1\]. This form reveals properties such as the center (\(h,k)\) and the orientation of the hyperbola.