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Find the equation of the given central conic. Hyperbola with a vertex at (0,-4) and a focus at (0,-5)

Short Answer

Expert verified
The equation is \((y+4.5)^2 = 0.25\), representing two lines \(y = -4\) and \(y = -5\).

Step by step solution

01

Identify the Type of Hyperbola

Since the vertex and focus are both on the y-axis and differing in y-value, the hyperbola is vertical. Thus, the equation takes the form of \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).
02

Locate the Center

For this vertical hyperbola, the center is midway between the vertex at \((0, -4)\) and the focus at \((0, -5)\). To find it, calculate the average: \( k = \frac{-4 + (-5)}{2} = -4.5 \), giving us the center at \((0, -4.5)\).
03

Calculate Distance to Vertex (a-value)

The distance between the vertex \((0, -4)\) and the center \((0, -4.5)\) gives us \(a\). Calculating gives \(a = |-4 - (-4.5)| = 0.5\).
04

Calculate Distance to Focus (c-value)

The distance between the focus \((0, -5)\) and the center \((0, -4.5)\) gives us \(c\). Calculating gives \(c = |-5 - (-4.5)| = 0.5\).
05

Use Relationship b^2 = c^2 - a^2

For hyperbolas, the formula \(c^2 = a^2 + b^2\) is used to find \(b\). We have \(c^2 = 0.5^2 = 0.25\) and \(a^2 = 0.5^2 = 0.25\), thus \(b^2 = c^2 - a^2 = 0.25 - 0.25 = 0\).
06

Write the Hyperbola Equation

Using the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), and substituting \(k = -4.5\), \(h = 0\), \(a = 0.5\), \(b = 0\) (making the denominator undefined for \(b\)), the equation simplifies to \((y+4.5)^2 = 0.25\). This represents a degenerate hyperbola, which is just two separate lines at \(y = -4\) and \(y = -5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conic Sections
Conic sections are a fascinating and fundamental concept in mathematics. They arise from intersecting a plane with a cone, resulting in different shapes such as circles, ellipses, parabolas, and hyperbolas. Each type of conic section has unique properties and equations that define them.

  • Circles and Ellipses: Formed when the plane cuts through the cone at an angle perpendicular to its axis or at a lesser angle.
  • Parabolas: Occur when the plane is parallel to one of the cone's slopes.
  • Hyperbolas: Created when the plane intersects both halves of the double cone.
Understanding these sections is vital because they model real-world phenomena, from planetary orbits (ellipses) to the paths of projectiles (parabolas). Hyperbolas, in particular, appear in navigation systems and the design of lenses.
Equation of a Hyperbola
The equation of a hyperbola depends on its orientation, whether it opens horizontally or vertically. A hyperbola is characterized by two branches that mirror each other and open in opposite directions. This distinct feature is described by its standard equations:

1. **Horizontal Hyperbola:** The equation is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Here, the center is at \((h, k)\), and the hyperbola opens left and right. 2. **Vertical Hyperbola:** The equation is \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). In this case, the center is still \((h, k)\), but the hyperbola opens up and down.

The parameters \(a\) and \(b\) determine the shapes of the hyperbola. Specifically:
  • \(a\) is the distance from the center to the vertices along the transverse axis.
  • \(b\) is derived from the relationship \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to the foci.
Vertical Hyperbola
A vertical hyperbola is a type of hyperbola where the branches open vertically. This orientation is essential to note because it affects the position of the vertices and foci. In a vertical hyperbola, the vertices lie along the y-axis if the center is at the origin, or more generally, along the line passing through the center parallel to the y-axis.

  • **Vertices:** Located at \((h, k \pm a)\).
  • **Foci:** Located at \((h, k \pm c)\).
  • **Asymptotes:** The asymptotes are lines that the hyperbola approaches but never touches. They intersect at the center and are given by the equations \(y = k \pm \frac{a}{b}(x-h)\).
The identification of a vertical hyperbola starts with the orientation of the vertices and foci. Using the coordinates provided, you can calculate the center and distances to obtain the correct equation. Moreover, observing how the vertical branches behave can help in understanding the dynamics of real-world systems and graphs of functions that model such behaviors.

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Most popular questions from this chapter

Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(\frac{-x^{2}}{9}+\frac{y^{2}}{4}=1\)

Find the equation of the given central conic. Hyperbola whose asymptotes are \(x \pm 2 y=0\) and that goes through the point (4,3)

The region bounded by the hyperbola $$ b^{2} x^{2}-a^{2} y^{2}=a^{2} b^{2} $$ and a vertical line through a focus is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Consider a bridge deck weighing \(\delta\) pounds per linear foot and supported by a cable, which is assumed to be of negligible weight compared to the bridge deck. The cable section \(O P\) from the lowest point (the origin) to a general point \(P(x, y)\) is shown in Figure \(13 .\) The forces acting on this section of cable are \(H=\) horizontal tension pulling at \(O\) \(T=\) tangential tension pulling at \(P\) \(W=\delta x=\) weight of \(x\) feet of bridge deck For equilibrium, the horizontal and vertical components of \(T\) must balance \(H\) and \(W,\) respectively. Thus, $$ \frac{T \sin \phi}{T \cos \phi}=\tan \phi=\frac{\delta x}{H} $$ That is, $$ \frac{d y}{d x}=\frac{\delta x}{H}, \quad y(0)=0 $$ Solve this differential equation to show that the cable hangs in the shape of a parabola. (Compare this result with that for the unloaded hanging cable of Problem 53 of Section \(6.9 .)\)

A chord of a parabola that is perpendicular to the axis and 1 unit from the vertex has length 1 unit. How far is it from the vertex to the focus?

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