Ellipses: Standard Form: Videos & Practice Problems

An ellipse is formed by slicing a cone with a plane at a slight angle, resulting in a shape defined by two axes: the semi-major axis (a) and the semi-minor axis (b). The equations for horizontal and vertical ellipses differ based on the orientation, represented as $x^2/a^2+y^2/b^2=1$ for horizontal and $x^2/b^2+y^2/a^2=1$ for vertical. The vertices and foci are critical points along the major axis, calculated using $c^2=a^2-b^2$.

Graph Ellipses at Origin

In this discussion, we explore the concept of ellipses, which are formed when a three-dimensional cone is intersected by a two-dimensional plane at a slight angle. This results in a shape that is more complex than a circle, requiring us to consider additional parameters for its graph and equation. Despite these complexities, there are notable similarities between circles and ellipses that can simplify our understanding.

To begin, recall that a circle is defined by its radius, which is the distance from the center to any point on the circle. The standard equation for a circle is given by:

x^2 + y^2 = r^2

where r represents the radius. When we stretch a circle horizontally, we create an ellipse characterized by two distinct distances: the semi-major axis and the semi-minor axis. The semi-major axis, denoted as a, is the longer axis, while the semi-minor axis, denoted as b, is the shorter one. For a horizontally stretched ellipse, the equation takes the form:

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

In this case, a is the distance along the x-axis, and b is the distance along the y-axis.

Conversely, when the circle is stretched vertically, the major axis aligns with the y-axis. The equation for a vertically stretched ellipse is modified accordingly:

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

Here, a is now associated with the y-axis, while b corresponds to the x-axis.

It is essential to remember that in both equations, a always represents the longest distance from the center of the ellipse to its edge. To relate the ellipse equations back to the circle, one can rewrite the circle's equation by dividing through by r^2, yielding:

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

This form closely resembles the equations for ellipses, highlighting the differences in symmetry: circles maintain uniformity with a single radius, while ellipses require two distinct distances, a and b, due to their varying shapes.

Understanding these concepts provides a solid foundation for graphing ellipses centered at the origin, allowing for further exploration of their properties and applications in mathematics.

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 key 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 major axis, which runs horizontally. To find the vertices, you need the distance a, known as the semi-major axis. The vertices can be calculated by moving a units to the right and left from the center, resulting in coordinates at (a, 0) and (-a, 0).

The foci, on the other hand, are determined using the relationship between the semi-major axis and the semi-minor axis. The distance to the foci, denoted as c, can be calculated using the equation:

$$c^2 = a^2 - b^2$$

where b is the semi-minor axis. The foci are then located at (c, 0) and (-c, 0) for a horizontal ellipse.

For example, if the equation of the ellipse reveals that a squared is 25, then a equals 5. The vertices would be at (5, 0) and (-5, 0). If b squared is 9, then b equals 3. Using the equation for c, we find:

$$c^2 = 25 - 9 = 16$$

Thus, c equals 4, placing the foci 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 graphing ellipses, as it affects the orientation and placement of these key points.

In summary, the vertices and foci of an ellipse provide important insights into its shape and symmetry. By understanding how to calculate these points using the semi-major and semi-minor axes, one can effectively analyze and graph ellipses in both horizontal and vertical 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

When dealing with ellipses that are not centered at the origin, the equation takes the form \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\), where \((h, k)\) represents the center of the ellipse. Here, \(h\) indicates the horizontal shift and \(k\) indicates the vertical shift. This structure is similar to function transformations, where shifts in the graph occur based on the values of \(h\) and \(k\).

To graph an ellipse, the first step is to identify the lengths of the major and minor axes. The major axis is associated with the larger denominator in the equation. For example, if the equation yields \(a^2 = 64\) and \(b^2 = 9\), then \(a\) (the semi-major axis) is \(\sqrt{64} = 8\) and \(b\) (the semi-minor axis) is \(\sqrt{9} = 3\).

Next, determine the orientation of the ellipse. If the larger number (in this case, \(64\)) is under the \(y\) term, the ellipse is vertical. Conversely, if it were under the \(x\) term, the ellipse would be horizontal.

Finding the center of the ellipse involves identifying \(h\) and \(k\) from the equation. For instance, if the equation is \((x-4)^2/9 + (y-2)^2/64 = 1\), then \(h = 4\) and \(k = 2\), placing the center at the point \((4, 2)\).

To locate the vertices of a vertical ellipse, add and subtract the value of \(a\) from \(k\). This means moving up and down from the center. For example, from the center \((4, 2)\), moving up \(8\) units gives the vertex at \((4, 10)\), and moving down \(8\) units gives the vertex at \((4, -6)\).

For the \(b\) points, which are horizontal in a vertical ellipse, add and subtract \(b\) from \(h\). Starting from the center \((4, 2)\), moving left \(3\) units results in the point \((1, 2)\), and moving right \(3\) units results in the point \((7, 2)\).

Finally, to graph the ellipse, connect the vertices and \(b\) points with a smooth curve. Additionally, the foci can be found using the formula \(c = \sqrt{a^2 - b^2}\). For a vertical ellipse, the foci are located at \((h, k \pm c)\). This means you would add and subtract \(c\) from the vertical position \(k\) to find the foci's coordinates.

Understanding these steps allows for the effective graphing and analysis of ellipses that are not centered at the origin, reinforcing the connection between algebraic equations and their graphical representations.

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)

Here’s what students ask on this topic:

What is the standard form equation of an ellipse?

The standard form equation of an ellipse depends on its orientation. For a horizontal ellipse, the equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. For a vertical ellipse, the equation is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$. Here, $a$ is the semi-major axis and $b$ is the semi-minor axis. These equations help define the shape and symmetry of the ellipse.

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

To find the vertices of an ellipse, identify the semi-major axis $a$. For a horizontal ellipse, the vertices are at $(a,0)$ and $(-a,0)$. For a vertical ellipse, they are at $(0,a)$ and $(0,-a)$. To find the foci, use the equation $c^2=a^2-b^2$. The foci are located at $(c,0)$ and $(-c,0)$ for horizontal ellipses, and $(0,c)$ and $(0,-c)$ for vertical ellipses.

How does the equation of an ellipse change when it is not centered at the origin?

When an ellipse is not centered at the origin, its equation includes shifts represented by $h$ and $k$. The equation for a horizontal ellipse becomes $\frac{{(}^x}{a^2}+\frac{{(}^y}{b^2}=1$. For a vertical ellipse, it is $\frac{{(}^x}{b^2}+\frac{{(}^y}{a^2}=1$. Here, $h$ and $k$ represent the horizontal and vertical shifts, respectively.

What is the relationship between the semi-major and semi-minor axes in an ellipse?

The semi-major axis, denoted as $a$, is the longest distance from the center of the ellipse to its perimeter, while the semi-minor axis, denoted as $b$, is the shortest. The relationship between these axes is crucial in defining the ellipse's shape. The equation $c^2=a^2-b^2$ helps calculate the distance to the foci, where $c$ is the distance from the center to each focus. This relationship is essential for understanding the symmetry and orientation of the ellipse.