Describe the hyperbola ( x + 2 ) 2 9 − ( y − 4 ) 2 16 = 1 \frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1 9 ( x + 2 ) 2 − 16 ( y − 4 ) 2 = 1 .

This is a horizontal hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 1 , 4 ) , ( − 5 , 4 ) \left(1,4\right),\left(-5,4\right) ( 1 , 4 ) , ( − 5 , 4 ) and foci at ( 3 , 4 ) , ( − 7 , 4 ) \left(3,4\right),\left(-7,4\right) ( 3 , 4 ) , ( − 7 , 4 ) .

This is a vertical hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 4 , 2 ) , ( 4 , − 6 ) \left(4,2\right),\left(4,-6\right) ( 4 , 2 ) , ( 4 , − 6 ) and foci at ( 4 , 4 ) , ( 4 , − 8 ) \left(4,4\right),\left(4,-8\right) ( 4 , 4 ) , ( 4 , − 8 ) .

This is a vertical hyperbola centered at ( 2 , − 4 ) \left(2,-4\right) ( 2 , − 4 ) with vertices at ( 4 , 1 ) , ( 4 , − 5 ) \left(4,1\right),\left(4,-5\right) ( 4 , 1 ) , ( 4 , − 5 ) and foci at ( 4 , 3 ) , ( 4 , − 7 ) \left(4,3\right),\left(4,-7\right) ( 4 , 3 ) , ( 4 , − 7 ) .

This is a horizontal hyperbola centered at ( − 2 , 4 ) \left(-2,4\right) ( − 2 , 4 ) with vertices at ( 2 , 4 ) , ( − 6 , 4 ) \left(2,4\right),\left(-6,4\right) ( 2 , 4 ) , ( − 6 , 4 ) and foci at ( 4 , 4 ) , ( − 8 , 4 ) \left(4,4\right),\left(-8,4\right) ( 4 , 4 ) , ( − 8 , 4 ) .

Describe the hyperbola y 2 − ( x − 1 ) 2 4 = 1 y^2-\frac{\left(x-1\right)^2}{4}=1 .

This is a vertical hyperbola centered at ( 1 , 0 ) \left(1,0\right) with vertices at ( 1 , 1 ) , ( 1 , − 1 ) \left(1,1\right),\left(1,-1\right) and foci at ( 1 , 5 ) , ( 1 , − 5 ) \left(1,\sqrt5\right),\left(1,-\sqrt5\right) .

This is a vertical hyperbola centered at ( 1 , 0 ) \left(1,0\right) with vertices at ( 1 , 2 ) , ( 1 , − 2 ) \left(1,2\right),\left(1,-2\right) and foci at ( 1 , 1 ) , ( 1 , − 1 ) \left(1,1\right),\left(1,-1\right) .

This is a horizontal hyperbola centered at ( − 1 , 0 ) \left(-1,0\right) with vertices at ( 0 , 0 ) , ( − 2 , 0 ) \left(0,0\right),\left(-2,0\right) and foci at ( 5 − 1 , 0 ) , ( − 5 − 1 , 0 ) \left(\sqrt5-1,0\right),\left(-\sqrt5-1,0\right) .

This is a horizontal hyperbola centered at ( 1 , 0 ) \left(1,0\right) with vertices at ( 0 , 0 ) , ( − 2 , 0 ) \left(0,0\right),\left(-2,0\right) and foci at ( 1 , 5 ) , ( 1 , − 5 ) \left(1,\sqrt5\right),\left(1,-\sqrt5\right) .

8. Conic Sections

Hyperbolas NOT at the Origin

8. Conic Sections

Hyperbolas NOT at the Origin: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding hyperbolas involves recognizing their equations and how they change when centered at points other than the origin. The standard equation for a hyperbola centered at the origin is modified by subtracting constants h and k, which represent horizontal and vertical shifts. For a vertical hyperbola, the vertices are determined by the value of a, while the b points are found using the value of b. The asymptotes are drawn through a box connecting these points, and the foci are calculated using the relationship $c^2 = a^2 + b^2$. This synthesis of concepts is crucial for graphing and analyzing hyperbolas effectively.

concept

Graph Hyperbolas NOT at the Origin

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Graph Hyperbolas NOT at the Origin Video Summary

Conic sections are a fundamental topic in geometry, encompassing various shapes such as circles, ellipses, parabolas, and hyperbolas. This summary focuses on hyperbolas, particularly those that are not centered at the origin. Understanding how to graph and analyze these hyperbolas involves recognizing the changes in their equations and applying transformation concepts.

The standard equation for a hyperbola centered at the origin is given by:

For a horizontal hyperbola: $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$

For a vertical hyperbola: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

When the hyperbola is shifted to a new center $(h, k)$, the equations adjust accordingly, where $h$ represents the horizontal shift and $k$ the vertical shift. This transformation mirrors the adjustments made for ellipses and other function transformations.

To graph a hyperbola not centered at the origin, follow these steps:

Identify whether the hyperbola is horizontal or vertical by examining the equation. The first term indicates the orientation; if it is positive, the hyperbola is vertical.
Determine the center $(h, k)$ from the equation. The values of $h$ and $k$ are derived from the constants subtracted from $x$ and $y$ respectively.
Find the vertices, which are located $a$ units away from the center along the axis of the hyperbola. Here, $a$ is the square root of the first positive term in the equation.
Calculate the $b$ points, which are located $b$ units away from the center along the transverse axis. The value of $b$ is derived from the second term in the equation.
Draw a box connecting the vertices and $b$ points to assist in finding the asymptotes. The asymptotes are lines that pass through the corners of this box.
Sketch the branches of the hyperbola, which approach the asymptotes but never intersect them.
Finally, determine the foci, which are located along the transverse axis. The distance $c$ from the center to each focus is calculated using the relationship $c^2 = a^2 + b^2$.

For example, if given a vertical hyperbola with the equation:

$\frac{(x - 2)^2}{9} - \frac{(y - 1)^2}{16} = 1$

Here, $h = 2$ and $k = 1$. The vertices would be at $(2, 4)$ and $(2, -2)$, while the $b$ points would be at $(6, 1)$ and $(-2, 1)$. The asymptotes can be drawn through the corners of the box formed by these points, and the foci can be calculated as $(2, 6)$ and $(2, -4)$ after determining $c$.

Understanding these steps allows for effective graphing and analysis of hyperbolas, enhancing comprehension of conic sections as a whole.

Problem

Describe the hyperbola $\frac{\left(x+2\right)^2}{9}-\frac{\left(y-4\right)^2}{16}=1$ .

This is a vertical hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(4,2\right),\left(4,-6\right)$ and foci at $\left(4,4\right),\left(4,-8\right)$ .

This is a vertical hyperbola centered at $\left(2,-4\right)$ with vertices at $\left(4,1\right),\left(4,-5\right)$ and foci at $\left(4,3\right),\left(4,-7\right)$ .

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(2,4\right),\left(-6,4\right)$ and foci at $\left(4,4\right),\left(-8,4\right)$ .

This is a horizontal hyperbola centered at $\left(-2,4\right)$ with vertices at $\left(1,4\right),\left(-5,4\right)$ and foci at $\left(3,4\right),\left(-7,4\right)$ .

Problem

Describe the hyperbola $y^{2} - \frac{{(x - 1)}^{2}}{4} = 1$ .

This is a vertical hyperbola centered at $open paren 1 comma 0 close paren$ with vertices at $open paren 1 comma 1 close paren comma open paren 1 comma negative 1 close paren$ and foci at $(1, \sqrt{5}), (1, - \sqrt{5})$ .

This is a vertical hyperbola centered at $open paren 1 comma 0 close paren$ with vertices at $open paren 1 comma 2 close paren comma open paren 1 comma negative 2 close paren$ and foci at $open paren 1 comma 1 close paren comma open paren 1 comma negative 1 close paren$ .

This is a horizontal hyperbola centered at $open paren negative 1 comma 0 close paren$ with vertices at $open paren 0 comma 0 close paren comma open paren negative 2 comma 0 close paren$ and foci at $(\sqrt{5} - 1, 0), (- \sqrt{5} - 1, 0)$ .

This is a horizontal hyperbola centered at $open paren 1 comma 0 close paren$ with vertices at $open paren 0 comma 0 close paren comma open paren negative 2 comma 0 close paren$ and foci at $(1, \sqrt{5}), (1, - \sqrt{5})$ .

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Here’s what students ask on this topic:

What is the standard equation of a hyperbola not centered at the origin?

The standard equation of a hyperbola not centered at the origin depends on its orientation. For a horizontal hyperbola, the equation is:

$\frac{(^{x - h) 2}}{a} - \frac{(^{y - k) 2}}{b} = 1$

For a vertical hyperbola, the equation is:

$\frac{(^{y - k) 2}}{a} - \frac{(^{x - h) 2}}{b} = 1$

Here, (h, k) represents the center of the hyperbola, a determines the distance to the vertices, and b determines the distance to the co-vertices.

How do you determine the center of a hyperbola not at the origin?

The center of a hyperbola not at the origin is determined by the values subtracted from x and y in the equation. For example, in the equation:

$\frac{(^{x - h) 2}}{a} - \frac{(^{y - k) 2}}{b} = 1$

The center is at (h, k). The value of h is the number subtracted from x, and k is the number subtracted from y. If the equation is:

$\frac{(^{y - k) 2}}{a} - \frac{(^{x - h) 2}}{b} = 1$

The center is still (h, k), but the hyperbola is vertically oriented.

How do you find the vertices of a hyperbola not centered at the origin?

To find the vertices of a hyperbola not centered at the origin, follow these steps:

1. Identify the center (h, k) from the equation.

2. Determine the value of a from the equation, where $a^{2}$ is the denominator of the first term.

3. For a horizontal hyperbola, the vertices are at:

$(h + a, k)$ and $(h - a, k)$ .

4. For a vertical hyperbola, the vertices are at:

$(h, k + a)$ and $(h, k - a)$ .

These points represent the farthest extent of the hyperbola along its major axis.

How do you calculate the asymptotes of a hyperbola not at the origin?

The asymptotes of a hyperbola are straight lines that pass through the center and guide the shape of the hyperbola. For a hyperbola centered at (h, k):

1. For a horizontal hyperbola:

$y - k = \pm \frac{b}{a} (x - h)$

2. For a vertical hyperbola:

$y - k = \pm \frac{a}{b} (x - h)$

Here, a and b are derived from the denominators of the equation. These equations represent the slopes of the asymptotes, and the lines pass through the center (h, k).