Conic Sections: Videos & Practice Problems

Conic sections include circles, ellipses, parabolas, and hyperbolas, each defined by unique equations and characteristics. A circle's equation is $x^2+y^2=r^2$. An ellipse has two axes, with its equation being $x^2/a^2+y^2/b^2=1$. Parabolas are defined by their focus and directrix, while hyperbolas feature asymptotes and vertices, with equations reflecting their orientation and center shifts.

Geometries from Conic Sections

Conic sections are geometric shapes formed by slicing a three-dimensional cone with a two-dimensional plane. Understanding these shapes involves recognizing their unique characteristics and equations, which can be derived from the way the cone is intersected by the plane.

The first conic section is the circle, which is created when the cone is sliced horizontally. This results in a perfectly round shape, and the standard equation for a circle centered at the origin is given by:

\(x^2 + y^2 = r^2\)

where \(r\) is the radius of the circle.

The second shape is the ellipse, which occurs when the cone is sliced at a slight angle. This results in an oval shape, resembling a stretched circle. The general equation for an ellipse centered at the origin is:

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

The parabola is formed when the plane is tilted at a steep angle, resulting in a U-shaped curve. The standard equation for a parabola that opens upwards is:

\(y = ax^2 + bx + c\)

where \(a\), \(b\), and \(c\) are constants that determine the shape and position of the parabola.

Finally, the hyperbola is created when the plane slices vertically through the cone, resulting in two separate curves that open away from each other. The standard equation for a hyperbola centered at the origin is:

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

where \(a\) and \(b\) define the distances from the center to the vertices and foci of the hyperbola.

Each of these conic sections has distinct properties and equations that are essential for solving related mathematical problems. By understanding how the shape is derived from the cone, students can better grasp the relationships between the different conic sections and their applications in various fields of study.

How can you slice a vertically oriented 3D cone to get a 2D parabola?

• A. Slice the cone with a horizontal plane.
• B. Slice the cone with a slightly tilted plane.
• C. Slice the cone with a heavily tilted plane.
• D. Slice the cone with a vertical plane.

Answer: (C)

How can you slice a vertically oriented 3D cone with a 2D plane to get a circle?

• A. Slice the cone with a horizontal plane.
• B. Slice the cone with a slightly tilted plane.
• C. Slice the cone with a heavily tilted plane.
• D. Slice the cone with a vertical plane.

Answer: (A)

A vertically oriented 3D cone is sliced with a vertical 2D plane. What is the conic section that will form?

• A. Circle:
• B. Ellipse:
• C. Parabola:
• D. Hyperbola:

Answer: (D)

Circles in Standard Form

Conic sections are fascinating shapes formed by slicing a three-dimensional cone with a two-dimensional plane. One of the most fundamental shapes in this category is the circle, which is created by slicing the cone horizontally. A circle is defined as the set of all points that are equidistant from a central point, known as the center. This unique property of circles allows us to describe their size and position using specific mathematical equations.

To graph a circle, two key components are required: the center and the radius. The center is typically denoted as (h, k), where h represents the horizontal position and k represents the vertical position. For example, if a circle is centered at the origin (0, 0), the equation of the circle can be expressed as:

\[x^2 + y^2 = r^2\]

Here, r is the radius of the circle. If the circle is not centered at the origin, the equation adjusts to account for the new center, taking the form:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

In this equation, h and k are the coordinates of the center, and r is the radius. For instance, if a circle has a center at (2, 1) and a radius of 4, the equation would be:

\[ (x - 2)^2 + (y - 1)^2 = 4^2 \]

When given the equation of a circle, one can derive its center and radius to graph it effectively. The center can be identified directly from the equation, while the radius can be determined by taking the square root of the constant on the right side of the equation. For example, if the equation is:

\[ (x - 1)^2 + (y - 2)^2 = 9 \]

the center is (1, 2) and the radius is 3, since \( r^2 = 9 \) implies \( r = \sqrt{9} = 3 \).

To graph the circle, one would plot the center and then mark points at a distance equal to the radius in all four cardinal directions (up, down, left, right) from the center. Connecting these points with a smooth curve results in the visual representation of the circle.

It is important to note that a circle does not represent a function in the mathematical sense, as it fails the vertical line test. This test states that if a vertical line intersects a graph at more than one point, the graph does not represent a function. Since a circle can be intersected by a vertical line at two points, it is classified as a non-function.

Understanding the properties and equations of circles is essential for further exploration of conic sections and their applications in various fields of study.

Sketch a graph of the circle based on the following equation: $x^2+\left(y-1\right)^2=9$

• A.
• B.
• C.
• D.

Answer: (D)

Sketch a graph of the circle based on the following equation: $\left(x-2\right)^2+\left(y+3\right)^2=1$

• A.
• B.
• C.
• D.

Answer: (B)

Find the equation for the following circle:

• A. $x^2+y^2=4$
• B. $\left(x+1\right)^2+y^2=4$
• C. $\left(x-1\right)^2+y^2=4$
• D. $\left(x+1\right)^2+y^2=16$

Answer: (D)

Circles in Standard Form Example 1

To write the equation of a circle, we start with the standard form, which is given by the equation:

\( (x - h)^2 + (y - k)^2 = r^2 \)

In this equation, \((h, k)\) represents the center of the circle, and \(r\) is the radius. Let's explore how to apply this formula through a few examples.

For the first circle, the center is at \((0, 1)\) and the radius is \(3\). Substituting these values into the standard form, we have:

\( (x - 0)^2 + (y - 1)^2 = 3^2 \)

This simplifies to:

\( x^2 + (y - 1)^2 = 9 \)

Next, consider a circle with its center at \((1, -2)\) and a radius of \(1\). Plugging these values into the standard form gives us:

\( (x - 1)^2 + (y + 2)^2 = 1^2 \)

This simplifies to:

\( (x - 1)^2 + (y + 2)^2 = 1 \)

For the final example, the center is at \((-6, 3)\) and the radius is \(\sqrt{5}\). Using the standard form, we write:

\( (x + 6)^2 + (y - 3)^2 = (\sqrt{5})^2 \)

This simplifies to:

\( (x + 6)^2 + (y - 3)^2 = 5 \)

In summary, to find the equation of a circle, identify the center and radius, and substitute these values into the standard form equation. This method allows you to graph the circle accurately based on its defined characteristics.

Circles in General Form

In the study of circles, understanding the general form of a circle's equation is crucial for converting it to standard form. The general form of a circle's equation can appear daunting, but with the right approach, it can be simplified. The general form is typically expressed as:

\[x^2 + y^2 + Dx + Ey + F = 0\]

To convert this equation into standard form, which is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle and \(r\) is the radius, one must follow a systematic process that involves completing the square.

The first step is to group the \(x\) and \(y\) terms together and move the constant term to the other side of the equation. For example, if we have:

\[x^2 + 2x + y^2 - 4y + 1 = 0\]

we would rearrange it to:

\[x^2 + 2x + y^2 - 4y = -1\]

Next, we complete the square for both the \(x\) and \(y\) terms. For the \(x\) terms, take the coefficient of \(x\) (which is 2), divide it by 2 to get 1, and then square it to obtain 1. We add this value inside the equation:

\[x^2 + 2x + 1\]

For the \(y\) terms, take the coefficient of \(y\) (which is -4), divide it by 2 to get -2, and square it to obtain 4. We add this value as well:

\[y^2 - 4y + 4\]

After adding these squares, we must also adjust the right side of the equation to maintain equality. Thus, we add 1 and 4 to the right side:

\[x^2 + 2x + 1 + y^2 - 4y + 4 = -1 + 1 + 4\]

This simplifies to:

\[ (x + 1)^2 + (y - 2)^2 = 4 \]

Now, we can identify the center of the circle as \((-1, 2)\) and the radius \(r\) as the square root of 4, which is 2.

To graph the circle, start at the center \((-1, 2)\) and move 2 units in all directions (up, down, left, right) to find the points on the circle. Connecting these points with a smooth curve will yield the graph of the circle.

This method of converting from general to standard form not only helps in understanding the properties of circles but also enhances graphing skills, making it easier to visualize the geometric representation of the equation.

Determine if the equation $x^2+y^2-2x+4y-4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c\left(0,0\right)$, radius $r=2$.
• B. Is a circle, center = $c\left(0,0\right)$, radius $r=3$ .
• C. Is a circle, center = $c\left(1,-2\right)$, radius $r=3$.
• D. Is not a circle.

Answer: (C)

Determine if the equation $x^3+y^2+4x-8y+4=0$ is a circle, and if it is, find its center and radius.

• A. Is a circle, center = $c\left(0,0\right)$, radius $r=4$.
• B. Is a circle, center = $c\left(0,0\right)$$c\left(2,-4\right)$ , radius $r=4$.
• C. Is a circle, center = $c\left(-2,4\right)$ $c\left(0,0\right)$, radius $r=4$.
• D. Is not a circle.

Answer: (D)

Graph Ellipses at Origin

In the study of conic sections, the ellipse is a fascinating shape that emerges when a three-dimensional cone is intersected by a two-dimensional plane at a slight angle. Unlike the circle, which is defined by a single radius, the ellipse requires the consideration of two distinct distances: the semi-major axis and the semi-minor axis. The semi-major axis, denoted as \( a \), represents the longest distance from the center of the ellipse to its edge, while the semi-minor axis, denoted as \( b \), is the shorter distance.

When visualizing a horizontal ellipse, the semi-major axis extends along the x-axis, and the semi-minor axis extends along the y-axis. The equation for a horizontal ellipse centered at the origin can be expressed as:

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

For example, if \( a = 4 \) and \( b = 3 \), the equation becomes:

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

Conversely, when the ellipse is vertically stretched, the semi-major axis aligns with the y-axis, and the semi-minor axis aligns with the x-axis. The equation for a vertical ellipse is similarly structured but with the positions of \( a \) and \( b \) switched:

\(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)

For instance, if \( a = 4 \) and \( b = 3 \), the equation for the vertical ellipse would be:

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

Understanding the relationship between the equations of circles and ellipses can also be insightful. The equation of a circle can be transformed into a form resembling that of an ellipse by dividing through by \( r^2 \), leading to:

\(\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1\)

This highlights the symmetry of the circle, where the radius is constant in all directions, contrasting with the varying distances in an ellipse.

In summary, the ellipse is characterized by its semi-major and semi-minor axes, which dictate its shape and orientation. The equations governing these ellipses provide a framework for understanding their geometric properties, making them a crucial topic in the study of conic sections.

Given the equation $\frac{x^2}{4}+\frac{y^2}{9}=1$, sketch a graph of the ellipse.

• A.
• B.
• C.
• D.

Answer: (B)

Given the ellipse equation $\frac{x^2}{16}+\frac{y^2}{4}=1$, determine the magnitude of the semi-major axis (a) and the semi-minor axis (b).

• A. $a=16$, $b=4$
• B. $a=4$, $b=16$
• C. $a=4$, $b=2$
• D. $a=2$, $b=4$

Answer: (C)

Foci and Vertices of an Ellipse

In the study of ellipses, understanding the concepts of vertices and foci is essential for graphing and analyzing their properties. An ellipse has two vertices and two foci, which are critical points located along the major axis of the ellipse. The vertices are the points that are furthest from the center of the ellipse, while the foci are points that help define the ellipse's symmetry.

For a horizontally stretched ellipse, the vertices are positioned along the x-axis, specifically at the coordinates \((a, 0)\) and \((-a, 0)\), where \(a\) represents the length of the semi-major axis. To find the foci, which are located at \((c, 0)\) and \((-c, 0)\), you need to calculate \(c\) using the relationship \(c^2 = a^2 - b^2\), where \(b\) is the length of the semi-minor axis. This equation is derived from the standard form of the ellipse equation, which is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

To illustrate, consider an ellipse with \(a^2 = 25\) and \(b^2 = 9\). First, calculate \(a\) by taking the square root of \(25\), resulting in \(a = 5\). The vertices are then located at \((5, 0)\) and \((-5, 0)\). Next, calculate \(b\) by taking the square root of \(9\), giving \(b = 3\). Using the equation for \(c\), we find \(c^2 = 25 - 9 = 16\), leading to \(c = 4\). Thus, the foci are positioned at \((4, 0)\) and \((-4, 0)\).

In contrast, for a vertically stretched ellipse, the vertices and foci are aligned along the y-axis. The vertices would be at \((0, a)\) and \((0, -a)\), while the foci would be at \((0, c)\) and \((0, -c)\). This distinction is crucial when determining the orientation of the ellipse based on its equation.

In summary, the vertices and foci of an ellipse provide valuable information about its shape and symmetry. By understanding how to calculate these points using the semi-major and semi-minor axes, students can effectively analyze and graph ellipses in various orientations.

Determine the vertices and foci of the following ellipse: $\frac{x^2}{49}+\frac{y^2}{36}=1$.

• A. Vertices: $\left(7,0\right),\left(-7,0\right)$Foci: $\left(6,0\right),\left(-6,0\right)$
• B. Vertices: $\left(6,0\right),\left(-6,0\right)$Foci: $\left(7,0\right),\left(-7,0\right)$
• C. Vertices: $\left(7,0\right),\left(-7,0\right)$Foci: $\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)$
• D. Vertices: $\left(0,7\right),\left(0,-7\right)$Foci: $\left(0,\sqrt{13}\right),\left(0,-\sqrt{13}\right)$

Answer: (C)

Determine the vertices and foci of the following ellipse: $\frac{x^2}{9}+\frac{y^2}{16}=1$.

• A. Vertices: $\left(4,0\right),\left(-4,0\right)$Foci: $\left(\sqrt7,0\right),\left(-\sqrt7,0\right)$
• B. Vertices: $\left(0,4\right),\left(0,-4\right)$Foci: $\left(0,\sqrt7\right),\left(0,-\sqrt7\right)$
• C. Vertices: $\left(4,0\right),\left(-4,0\right)$Foci: $\left(3,0\right),\left(-3,0\right)$
• D. Vertices: $\left(0,4\right),\left(0,-4\right)$Foci: $\left(0,3\right),\left(0,-3\right)$

Answer: (B)

Find the standard form of the equation for an ellipse with the following conditions.
Foci = $\left(-5,0\right),\left(5,0\right)$
Vertices = $\left(-8,0\right),\left(8,0\right)$

• A. $\frac{x^2}{64}+\frac{y^2}{25}=1$
• B. $\frac{x^2}{25}+\frac{y^2}{64}=1$
• C. $\frac{x^2}{8}+\frac{y^2}{5}=1$
• D. $\frac{x^2}{64}+\frac{y^2}{39}=1$

Answer: (D)

Graph Ellipses NOT at Origin

In studying ellipses, it's essential to understand how their equations change when they are not centered at the origin. When an ellipse is shifted to a new location defined by the coordinates (h, k), the general equation takes the form:

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

Here, h represents the horizontal shift, and k represents the vertical shift. The values a and b correspond to the lengths of the semi-major and semi-minor axes, respectively. The major axis is associated with the larger denominator, while the minor axis is linked to the smaller denominator.

To graph an ellipse not centered at the origin, follow these steps:

1. **Identify the Major and Minor Axes**: Determine a and b by identifying the larger and smaller numbers in the denominators of the equation. For example, if the equation shows that \(a^2 = 64\) and \(b^2 = 9\), then \(a = 8\) and \(b = 3\).

2. **Determine Orientation**: The orientation of the ellipse (vertical or horizontal) is determined by the position of the larger number. If \(a^2\) is under the y-term, the ellipse is vertical; if it is under the x-term, it is horizontal.

3. **Find the Center**: The center of the ellipse is located at the point (h, k). From the equation, h is found by looking at the term subtracted from x, and k is found from the term subtracted from y.

4. **Locate the Vertices**: For a vertical ellipse, add and subtract the value of a from k to find the vertices. For instance, if k = 2 and a = 8, the vertices would be at (h, k + a) and (h, k - a), resulting in points at (4, 10) and (4, -6).

5. **Find the B Points**: For vertical ellipses, the b points are found by adding and subtracting b from h. If b = 3, the b points would be at (h - b, k) and (h + b, k), resulting in points at (1, 2) and (7, 2).

6. **Draw the Ellipse**: Connect the vertices and b points with a smooth curve to complete the graph of the ellipse.

Additionally, to find the foci of the ellipse, calculate the value of c using the formula:

\[c = \sqrt{a^2 - b^2}\]

For a vertical ellipse, the foci are located at (h, k ± c). This means you will add and subtract c from the vertical position k to find the foci's coordinates.

Understanding these steps allows for a comprehensive approach to graphing and analyzing ellipses that are not centered at the origin, reinforcing the connection between transformations and conic sections.

Graph the ellipse $\frac{\left(x-1\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$.

• A.
• B.
• C.
• D.

Answer: (A)

Determine the vertices and foci of the ellipse $\left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1$.

• A. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right)$
• B. Vertices: $\left(-1,4\right),\left(-1,0\right)$Foci: $\left(-2,2\right),\left(0,2\right)$
• C. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(1,2+\sqrt3\right),\left(1,2-\sqrt3\right)$
• D. Vertices: $\left(-2,2\right),\left(0,2\right)$Foci: $\left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right)$

Answer: (A)

Parabolas as Conic Sections

In the study of conic sections, the parabola is a significant shape that can be visualized by slicing a three-dimensional cone with a tilted plane. Understanding parabolas involves identifying two key components: the focus and the directrix. The focus is a specific point on the graph, while the directrix is a line. Both the focus and the directrix are equidistant from the vertex of the parabola, although they may be oriented in different directions.

When analyzing a parabola, the orientation determines the location of the focus and directrix. For a parabola that opens upwards, the focus is positioned above the vertex, while the directrix is below it. Conversely, if the parabola opens downwards, the focus is below the vertex, and the directrix is above. The relationship between the vertex, focus, and directrix is governed by the parameter p, which represents the distance from the vertex to both the focus and the directrix.

The standard form of a parabola's equation is given by:

\[y = \frac{1}{4p}(x - h)^2 + k\]

where \((h, k)\) is the vertex of the parabola. In this equation, the term \(4p\) is crucial for determining the shape and position of the parabola. For example, if the equation is simplified to \(y = 4x\), we can set \(4p = 4\) to find that \(p = 1\). This indicates that the focus is located at \((0, 1)\) and the directrix is the line \(y = -1\).

To graph a parabola, start by identifying the vertex from the equation. For instance, if the vertex is at \((1, 2)\), the next step is to determine the value of \(p\). If \(4p = 8\), then \(p = 2\). This means the focus will be at \((1, 4)\) and the directrix will be the line \(y = 0\). The width of the parabola can be established by moving \(2p\) units horizontally from the focus, yielding points that help define the parabola's shape.

Finally, connecting these points with a smooth curve will yield the parabola, and the directrix can be drawn accordingly. Understanding these relationships and steps is essential for effectively working with parabolas in conic sections.

Graph the parabola $-4\left(y+1\right)=\left(x+1\right)^2$, and find the focus point and directrix line.

• A.
• B.
• C.
• D.

Answer: (D)

If a parabola has the focus at $\left(0,-1\right)$ and a directrix line $y=1$, find the standard equation for the parabola.

• A. $4y=x^2$
• B. $4\left(y-1\right)=x^2$
• C. $-4y=x^2$
• D. $-4\left(y+1\right)=x^2$

Answer: (C)

Parabolas as Conic Sections Example 1

To analyze parabolas, we utilize the standard equation for a parabola, which is given by:

\[4p(y - k) = (x - h)^2\]

In this equation, \((h, k)\) represents the vertex of the parabola. If \(h\) and \(k\) are absent, the parabola is centered at the origin, simplifying the equation to:

\[4py = x^2\]

When solving for the vertex, focus, and directrix, we can follow a systematic approach. For example, consider a parabola where the equation is in the form \(4py = x^2\). Here, the vertex is at the origin \((0, 0)\). To find the focus and directrix, we set \(4p\) equal to the coefficient in front of \(y\). If \(4p = 16\), then:

\[p = \frac{16}{4} = 4\]

Since \(p\) is positive, the parabola opens upward. The focus, therefore, is located at \((0, 4)\), and the directrix is the line \(y = -4\).

In another example, if we have \(4p = \frac{1}{3}\), we can find \(p\) by rearranging:

\[p = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}\]

Again, since \(p\) is positive, the parabola opens upward, placing the focus at \((0, \frac{1}{12})\) and the directrix at \(y = -\frac{1}{12}\).

When the parabola is not centered at the origin, such as in the case where \(h\) and \(k\) are present, we identify the vertex directly from the equation. For instance, if the vertex is at \((2, 1)\), we can find \(p\) by setting \(4p\) equal to the coefficient in front of \(y - k\). If \(4p = 8\), then:

\[p = \frac{8}{4} = 2\]

With \(p\) being positive, the focus is at \((2, 3)\) and the directrix is the line \(y = -1\).

In cases where the equation indicates a downward-opening parabola, such as \(4p = -12\), we find:

\[p = \frac{-12}{4} = -3\]

Here, the negative value of \(p\) indicates that the parabola opens downward. The vertex might be at \((-1, 0)\), leading to a focus at \((-1, -3)\) and a directrix at \(y = 3\).

In summary, by identifying the vertex, calculating \(p\), and determining the focus and directrix based on the sign of \(p\), we can effectively analyze and understand the properties of parabolas without needing a graph. This methodical approach allows for clear identification of key features in various forms of parabolic equations.

Horizontal Parabolas

In the study of parabolas, we typically encounter vertical parabolas that open either upwards or downwards. However, when we consider horizontal parabolas, the equations and characteristics change slightly, primarily involving the switching of the x and y variables. This transformation leads to a new understanding of how these parabolas behave and how to graph them effectively.

For horizontal parabolas, the general form of the equation is derived from the vertical parabola's equation by interchanging the roles of x and y. For instance, while a vertical parabola can be expressed as \(4p y = x^2\), the corresponding horizontal parabola is represented as \(4p x = y^2\). This switch indicates that the axis of symmetry for horizontal parabolas is horizontal, and the directrix, which is always perpendicular to the axis of symmetry, becomes vertical.

When analyzing the orientation of the parabola, the value of \(p\) plays a crucial role. A positive \(p\) value indicates that a vertical parabola opens upwards, while a negative \(p\) value means it opens downwards. Conversely, for horizontal parabolas, a positive \(p\) indicates the parabola opens to the right, and a negative \(p\) indicates it opens to the left. This relationship can be visualized on a graph, where positive values typically extend upwards and to the right, while negative values extend downwards and to the left.

To graph a horizontal parabola, one must follow a systematic approach. First, identify the vertex, which is often at the origin if no \(h\) or \(k\) values are present. Next, calculate the \(p\) value from the equation, recognizing that \(4p\) corresponds to the coefficient of \(x\) or \(y\). For example, if the equation is \(8x = y^2\), then \(4p = 8\) leads to \(p = 2\).

Once \(p\) is determined, locate the focus of the parabola, which lies \(p\) units in the direction the parabola opens. For a parabola that opens to the right, the focus would be at \((2, 0)\) if starting from the origin. The width of the parabola can be established by moving \(2p\) units vertically from the focus, resulting in points at \((2, 4)\) and \((2, -4)\). Finally, connect these points with a smooth curve to complete the graph.

Lastly, the directrix is found by moving \(p\) units in the opposite direction of the opening. For a parabola that opens to the right, the directrix would be the line \(x = -2\). This comprehensive understanding of horizontal parabolas, including their equations, graphing techniques, and properties, allows for effective problem-solving and visualization in mathematics.

Graph the parabola $8\left(x+1\right)=\left(y-2\right)^2$ , and find the focus point and directrix line.

• A.
• B.
• C.
• D.

Answer: (B)

If a parabola has the focus at $\left(2,4\right)$ and a directrix line $x=-4$ , find the standard equation for the parabola.

• A. $12\left(x+1\right)=\left(y-4\right)^2$
• B. $-\left(x+1\right)=\left(y-4\right)^2$
• C. $12x=y^2$
• D. $4\left(x-1\right)=\left(y+4\right)^2$

Answer: (A)

Horizontal Parabolas Example 1

In studying parabolas, it's essential to understand their geometric properties, including the vertex, focus, and directrix. A horizontal parabola can be represented by the equation:

\(4p(x - h) = (y - k)^2\)

Here, \((h, k)\) denotes the vertex, and \(p\) represents the distance from the vertex to the focus and the directrix. A positive \(p\) indicates that the parabola opens to the right, while a negative \(p\) indicates it opens to the left.

For example, consider a horizontal parabola where the equation is given without any terms subtracted from \(x\) or \(y\). This indicates that the vertex is at the origin \((0, 0)\). If we set \(4p\) equal to a constant from the equation, we can solve for \(p\). If \(4p = 4\), then \(p = 1\). Thus, the focus is located at \((1, 0)\) and the directrix is the line \(x = -1\).

In another case, if the equation shows a subtraction from \(y\), such as \(y - 2\), the vertex shifts to \((0, 2)\). Setting \(4p\) equal to a different constant, say \(9\), gives \(p = \frac{9}{4}\). The focus would then be at \((\frac{9}{4}, 2)\) and the directrix at \(x = -\frac{9}{4}\).

When the equation includes terms like \(y + 2\), it’s crucial to recognize that this translates to \(y - (-2)\), indicating that \(k = -2\). If \(4p = 16\), then \(p = 4\), placing the focus at \((8, -2)\) and the directrix at \(x = 0\).

Lastly, if the equation indicates a subtraction from \(x\) (e.g., \(x - 1\)), the vertex is at \((1, 0)\). If \(4p = -2\), then \(p = -\frac{1}{2}\), indicating the parabola opens to the left. The focus would be at \((\frac{1}{2}, 0)\) and the directrix at \(x = \frac{3}{2}\).

Understanding these relationships allows for the effective identification of the vertex, focus, and directrix of horizontal parabolas, which is fundamental in the study of conic sections.

Introduction to Hyperbolas

In the study of conic sections, hyperbolas present a unique challenge due to their distinct characteristics and equations. The fundamental equation for a hyperbola closely resembles that of an ellipse, with the key difference being the presence of a minus sign instead of a plus sign. For a horizontal hyperbola, the equation can be expressed as:

\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]

Here, the value of a represents the distance from the center of the hyperbola to each of its two branches, while b plays a crucial role in graphing the hyperbola, although its interpretation differs from that in ellipses. In contrast, the equation for a vertical hyperbola is structured as:

\[\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1\]

In this case, a is still the distance to the curves, but it is oriented vertically, while b extends horizontally. It is important to note that in hyperbolas, the a value does not necessarily represent the largest number; rather, it is always the first term in the equation.

When graphing hyperbolas, identifying the orientation is essential. A hyperbola with a y term first indicates a vertical orientation, while an x term first indicates a horizontal orientation. The values of a and b are squared in the equations, and understanding their geometric implications is vital for accurate graphing.

To solidify these concepts, consider matching hyperbola equations to their corresponding graphs. For instance, if an equation starts with y squared, it suggests a vertical hyperbola, and the value of a can be determined by taking the square root of the first term. This process allows for the identification of the correct graph by checking the distances from the center to the curves.

In summary, mastering the equations and characteristics of hyperbolas is crucial for understanding their unique properties within the broader context of conic sections. With practice, the similarities and differences between hyperbolas and other conic shapes, such as ellipses, will become clearer, enhancing overall comprehension.

Given the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, find the length of the $a$-axis and the $b$-axis.

• A. $a=25,b=9$
• B. $a=9,b=25$
• C. $a=5,b=3$
• D. $a=3,b=5$

Answer: (C)

Given the hyperbola $x^2-\frac{y^2}{4}=1$, find the length of the $a$-axis and $b$-axis.

• A. $a=1,b=4$
• B. $a=4,b=1$
• C. $a=1,b=2$
• D. $a=2,b=1$

Answer: (C)

Given the hyperbola $\frac{y^2}{100}-\frac{x^2}{139}=1$, find the length of the $a$-axis and the $b$-axis.

• A. $a=100,b=139$
• B. $a=139,b=100$
• C. $a=\sqrt{139},b=10$
• D. $a=10,b=\sqrt{139}$

Answer: (D)

Foci and Vertices of Hyperbolas

In the study of hyperbolas, understanding how to find the vertices and foci is essential. A hyperbola, like an ellipse, has two vertices and two foci located along its major axis. The vertices are the points closest to the center of the hyperbola, and the distance from the center to each vertex is denoted as a.

To find the vertices, one must first identify the center of the hyperbola, which is often given in the standard form of the equation. For example, if the hyperbola is centered at the origin (0,0) and the equation is in the form:

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]

the value of a can be determined by taking the square root of the first denominator. If a² equals 4, then a is 2. The vertices can then be found by moving a units left and right from the center, resulting in the points (-2, 0) and (2, 0).

The foci of a hyperbola are unique points where the difference in distances from any point on the hyperbola to each focus remains constant. To calculate the foci, the distance c is needed, which can be derived from the equation:

\[c^2 = a^2 + b^2\]

Here, b² is the second term in the denominator of the hyperbola's equation. For instance, if b² equals 9, then b is 3. Using the previously calculated a² of 4, we find:

\[c^2 = 4 + 9 = 13\]

Taking the square root gives c approximately equal to 3.6. The foci are then located at (±√13, 0), which are approximately (3.6, 0) and (-3.6, 0).

It is important to note that the orientation of the hyperbola affects the placement of the vertices and foci. For a horizontal hyperbola, both the vertices and foci lie on the x-axis, while for a vertical hyperbola, they are positioned on the y-axis. In the case of a vertical hyperbola, the vertices would be at (0, ±a) and the foci at (0, ±c).

In summary, finding the vertices and foci of a hyperbola involves understanding the relationship between a, b, and c, and recognizing how the orientation of the hyperbola influences their positions.

Determine the vertices and foci of the hyperbola $\frac{y^2}{4}-x^2=1$.

• A. Vertices: $\left(2,0\right),\left(-2,0\right)$Foci: $\left(\sqrt5,0\right),\left(-\sqrt5,0\right)$
• B. Vertices: $\left(0,2\right),\left(0,-2\right)$Foci: $\left(0,\sqrt5\right),\left(0,-\sqrt5\right)$
• C. Vertices: $\left(1,0\right),\left(-1,0\right)$Foci: $\left(5,0\right),\left(-5,0\right)$
• D. Vertices: $\left(0,1\right),\left(0,-1\right)$Foci: $\left(0,5\right),\left(0,-5\right)$

Answer: (B)

Find the equation for a hyperbola with a center at $\left(0,0\right)$, focus at $\left(0,-6\right)$ and vertex at $\left(0,4\right)$ .

• A. $\frac{y^2}{16}-\frac{x^2}{20}=1$
• B. $\frac{y^2}{20}-\frac{x^2}{16}=1$
• C. $\frac{y^2}{4}-\frac{x^2}{\sqrt{20}}=1$
• D. $\frac{y^2}{\sqrt{20}}-\frac{x^2}{4}=1$

Answer: (A)

Asymptotes of Hyperbolas

Understanding the asymptotes of hyperbolas is crucial for graphing these shapes accurately. Asymptotes are lines that help define the behavior of hyperbolas as they extend towards infinity. To find the asymptotes, we utilize the values of a and b, which represent the distances from the center to the vertices and co-vertices, respectively. For a hyperbola in standard form, the relationship is given by a² and b² in the denominators of the equation.

To illustrate, consider a vertical hyperbola where a² = 9 and b² = 4. From these, we find a = 3 and b = 2. These values help us construct a rectangle (or box) that connects the vertices and co-vertices. The asymptotes can then be drawn through the corners of this box, extending through the center of the hyperbola.

The equations of the asymptotes for a vertical hyperbola can be expressed as:

y = ±\(\frac{a}{b}\)x

In our example, this results in the equations y = \(\frac{3}{2}\)x and y = -\(\frac{3}{2}\)x. This pattern simplifies the process of finding asymptotes significantly.

For horizontal hyperbolas, the approach is similar, but the roles of a and b are reversed. The equations for the asymptotes become:

y = ±\(\frac{b}{a}\)x

Using the same values, for a horizontal hyperbola, we would have the equations y = \(\frac{2}{3}\)x and y = -\(\frac{2}{3}\)x. This flipping of a and b is key to determining the correct asymptotes for horizontal hyperbolas.

In summary, the process of finding asymptotes involves identifying the values of a and b, constructing a box around the vertices and co-vertices, and then applying the appropriate formulas based on the orientation of the hyperbola. This systematic approach not only aids in graphing but also enhances understanding of hyperbolic behavior in mathematical contexts.

Find the equations for the asymptotes of the hyperbola $\frac{x^2}{64}-\frac{y^2}{100}=1$.

• A. $y=\pm\frac45x$
• B. $y=\pm\frac54x$
• C. $y=\pm\frac{16}{25}x$
• D. $y=\pm\frac{25}{16}x$

Answer: (B)

Find the equations for the asymptotes of the hyperbola $\frac{y^2}{16}-\frac{x^2}{9}=1$.

• A. $y=\pm\frac{9}{16}x$
• B. $y=\pm\frac{16}{9}x$
• C. $y=\pm\frac34x$
• D. $y=\pm\frac43x$

Answer: (D)

Graph Hyperbolas at the Origin

To graph a hyperbola from its equation, it is essential to first determine its orientation—whether it is horizontal or vertical. This can be identified by examining the equation; if the term with \(x^2\) appears first in the denominator, the hyperbola is horizontal. For example, if the equation is of the form \(\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\), it indicates a vertical hyperbola, while \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) indicates a horizontal hyperbola.

Next, identify the vertices of the hyperbola. For a horizontal hyperbola, the vertices are located at \((h \pm a, k)\), where \(a\) is derived from \(a^2\) in the equation. For instance, if \(a^2 = 9\), then \(a = 3\), placing the vertices at \((h + 3, k)\) and \((h - 3, k)\).

Following this, determine the \(b\) values, which are found from \(b^2\) in the equation. If \(b^2 = 64\), then \(b = 8\). The \(b\) points for a horizontal hyperbola are located at \((h, k \pm b)\), resulting in points at \((h, k + 8)\) and \((h, k - 8)\).

To find the asymptotes, draw a rectangle using the vertices and \(b\) points. The asymptotes are then represented by the lines that pass through the corners of this rectangle, following the equations \(y - k = \pm \frac{b}{a}(x - h)\).

Finally, sketch the branches of the hyperbola, which approach the asymptotes but never intersect them. The branches will open towards the vertices. To find the foci, use the relationship \(c^2 = a^2 + b^2\). For example, if \(a^2 = 9\) and \(b^2 = 64\), then \(c^2 = 9 + 64 = 73\), leading to \(c = \sqrt{73} \approx 8.54\). The foci are located at \((h \pm c, k)\), which for a horizontal hyperbola means they will be positioned along the x-axis.

In summary, the steps to graph a hyperbola from its equation include determining its orientation, finding the vertices and \(b\) points, drawing the asymptotes, sketching the branches, and locating the foci using the derived values of \(a\), \(b\), and \(c\).

Graphing Hyperbolas at the Origin Example 1

In this example, we explore the process of graphing a hyperbola, specifically focusing on a horizontal hyperbola. The orientation of the hyperbola is determined by the equation, where the presence of \(x^2\) first indicates that it opens along the x-axis. This leads us to identify the vertices, which are crucial points on the graph.

To find the vertices, we utilize the relationship between the equation and its components. The value of \(a^2\) is derived from the first term in the denominator, which is 16. Taking the square root gives us \(a = \sqrt{16} = 4\). Therefore, the vertices are located at the coordinates (4, 0) and (-4, 0).

Next, we calculate the b points, which are positioned perpendicular to the vertices. Here, \(b^2\) corresponds to the second term in the denominator, which is 20. The square root of 20 can be expressed as \(b = \sqrt{20} \approx 4.47\). Thus, the b points are at (0, \( \sqrt{20} \)) and (0, -\( \sqrt{20} \)), or approximately (0, 4.47) and (0, -4.47).

To visualize the hyperbola, we draw a box that connects the vertices and b points. The asymptotes, which guide the branches of the hyperbola, are then established by drawing lines through the corners of this box. The branches of the hyperbola extend from the vertices and approach the asymptotes, creating the characteristic shape of the hyperbola.

Finally, we determine the foci, which are located along the x-axis for a horizontal hyperbola. The relationship \(c^2 = a^2 + b^2\) is used, where \(a^2 = 16\) and \(b^2 = 20\). This results in \(c^2 = 36\), leading to \(c = \sqrt{36} = 6\). Consequently, the foci are positioned at (6, 0) and (-6, 0).

In summary, the graph of the hyperbola is constructed by identifying the vertices, b points, asymptotes, and foci, all of which are essential for accurately representing the hyperbola's structure and behavior.

Graph Hyperbolas NOT at the Origin

In the study of conic sections, hyperbolas are a fascinating shape that can be centered at various locations on a graph, not just at the origin. When dealing with a hyperbola that is shifted from the origin, the standard equation changes slightly. The general form of a hyperbola centered at the origin is given by:

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

When the hyperbola is centered at a point \((h, k)\), the equation becomes:

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

Here, \(h\) and \(k\) represent the horizontal and vertical shifts from the origin, respectively. The values \(a\) and \(b\) are derived from the equation, where \(a^2\) is associated with the positive term and \(b^2\) with the negative term.

To graph a hyperbola centered at a point other than the origin, follow these steps:

1. **Identify the Orientation**: Determine if the hyperbola is vertical or horizontal by examining the equation. If the positive term involves \(y\), it is vertical; if it involves \(x\), it is horizontal.

2. **Find the Center**: The center \((h, k)\) can be identified directly from the equation. For example, if the equation is \(\frac{(y - 1)^2}{9} - \frac{(x - 2)^2}{16} = 1\), then \(h = 2\) and \(k = 1\), placing the center at the point \((2, 1)\).

3. **Locate the Vertices**: For a vertical hyperbola, the vertices are found by adding and subtracting \(a\) from \(k\). If \(a^2 = 9\), then \(a = 3\). The vertices will be at \((h, k + a)\) and \((h, k - a)\), resulting in points \((2, 4)\) and \((2, -2)\).

4. **Determine the B Points**: The \(b\) values are found similarly, where \(b^2\) corresponds to the second term. If \(b^2 = 16\), then \(b = 4\). The \(b\) points are located at \((h + b, k)\) and \((h - b, k)\), giving points \((6, 1)\) and \((-2, 1)\).

5. **Draw the Asymptotes**: Construct a rectangle connecting the vertices and \(b\) points. The asymptotes are drawn through the corners of this rectangle, guiding the shape of the hyperbola.

6. **Sketch the Hyperbola**: The branches of the hyperbola will approach the asymptotes, starting from the vertices.

7. **Find the Foci**: The foci are determined using the relationship \(c^2 = a^2 + b^2\). For our example, \(c^2 = 9 + 16 = 25\), thus \(c = 5\). The foci are located at \((h, k + c)\) and \((h, k - c)\), resulting in points \((2, 6)\) and \((2, -4)\).

By following these steps, one can effectively graph a hyperbola centered at any point and understand its properties, including vertices, foci, and asymptotes. This process mirrors the transformations seen in other conic sections, making it a valuable skill in geometry.

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

• A. 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 $\left(4,4\right),\left(4,-8\right)$.
• B. 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 $\left(4,3\right),\left(4,-7\right)$.
• C. 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 $\left(4,4\right),\left(-8,4\right)$.
• D. 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 $\left(3,4\right),\left(-7,4\right)$.

Answer: (D)

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

• A. This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,1\right),\left(1,-1\right)$ and foci at $\left(1,\sqrt5\right),\left(1,-\sqrt5\right)$.
• B. This is a vertical hyperbola centered at $\left(1,0\right)$ with vertices at $\left(1,2\right),\left(1,-2\right)$ and foci at $\left(1,1\right),\left(1,-1\right)$.
• C. This is a horizontal hyperbola centered at $\left(-1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(\sqrt5-1,0\right),\left(-\sqrt5-1,0\right)$.
• D. This is a horizontal hyperbola centered at $\left(1,0\right)$ with vertices at $\left(0,0\right),\left(-2,0\right)$ and foci at $\left(1,\sqrt5\right),\left(1,-\sqrt5\right)$.

Answer: (A)

Here’s what students ask on this topic:

What are conic sections and how are they formed?

Conic sections are geometric shapes formed by slicing a three-dimensional cone with a two-dimensional plane. Depending on the angle and position of the slice, four distinct shapes can be created: circles, ellipses, parabolas, and hyperbolas. A horizontal slice produces a circle, while a slightly tilted slice forms an ellipse. A heavily tilted slice results in a parabola, and a vertical slice creates a hyperbola. Each conic section has a unique equation and set of characteristics that define its shape and properties. These shapes are widely studied in mathematics due to their applications in physics, engineering, and astronomy.

What is the standard equation of a circle and how do you graph it?

The standard equation of a circle is $x^2+y^2=r^2$, where $r$ is the radius of the circle. To graph a circle, identify its center and radius. If the center is at the origin (0,0), plot the center and mark points at a distance of $r$ units in all directions (up, down, left, and right). Connect these points with a smooth curve to form the circle. If the center is not at the origin, use the equation $(x\; -\; h)2^{}+\; (y\; -\; k)2^{}=r^2$, where $h$ and $k$ are the horizontal and vertical shifts of the center.

How do you find the vertices and foci of an ellipse?

To find the vertices and foci of an ellipse, start with its equation: $x^2/a^2+y^2/b^2=\; 1$ for a horizontal ellipse or $y^2/a^2+x^2/b^2=\; 1$ for a vertical ellipse. The vertices are located at a distance of $a$ from the center along the major axis. The foci are found using the formula $c^2=a^2-b^2$, where $c$ is the distance from the center to each focus along the major axis. For a horizontal ellipse, the vertices are at ($h$ ± $a$, $k$) and the foci are at ($h$ ± $c$, $k$). For a vertical ellipse, switch the roles of $x$ and $y$.

What is the difference between a hyperbola and an ellipse?

The primary difference between a hyperbola and an ellipse lies in their equations and shapes. An ellipse has the equation $x^2/a^2+y^2/b^2=\; 1$, with a plus sign, while a hyperbola has the equation $x^2/a^2-y^2/b^2=\; 1$, with a minus sign. Visually, an ellipse is a closed curve resembling an elongated circle, while a hyperbola consists of two open curves that diverge away from each other. The hyperbola also features asymptotes, which are diagonal lines that the curves approach but never touch. In contrast, an ellipse has no asymptotes and is bounded.

How do you graph a parabola using its focus and directrix?

To graph a parabola using its focus and directrix, start with its equation: $(x\; -\; h)2^{}=\; 4p(y\; -\; k)$ for a vertical parabola or $(y\; -\; k)2^{}=\; 4p(x\; -\; h)$ for a horizontal parabola. The vertex is at ($h$, $k$). The focus is located $p$ units from the vertex in the direction the parabola opens, while the directrix is a line $p$ units in the opposite direction. Plot the vertex, focus, and directrix, then sketch the parabola by ensuring all points are equidistant from the focus and directrix. For horizontal parabolas, switch the roles of $x$ and $y$.