9.04 Ellipses

The locus definition of ellipses

We have now seen a few conic sections including circles and parabolas. Now we will look at ellipses. Circles are just a special case of an ellipse.

The classic example of constructing an ellipse is to place two pins some distance apart. Tie each end of a string around each pin and then pull the string taut with a pencil. Then trace the curve that the string will allow, keeping it taut.

Try this in the applet below. The two pins are located at $F$F and $F'$F′. The pencil is located at the point $P$P.

Created with Geogebra

The shape created in this way is always an ellipse. The points located at the pins are the foci (or focus if singular).

Notice that the length of the string never changes. So the length of string from one focus to a point on the ellipse added to the length of string from the other focus to the same point will always equal a constant for that ellipse. The constant for the above ellipse is $4$4.

 

 

Worked example

Question 1

Consider the ellipse below.

Find the value of $x$x.

Think: A pair of lines meeting at a point on the ellipse from the foci will have the same total length. There are two pairs of these lines on this ellipse. The sum of one pair must be equal to the sum of the other pair.

Do: Construct an equation relating the length of the two pairs of lines and solve for $x$x.

$GP+FP$GP+FP	$=$=	$GQ+FQ$GQ+FQ
$x+x$x+x	$=$=	$9+3$9+3	(Substitution)
$2x$2x	$=$=	$12$12	(Collecting like terms)
$x$x	$=$=	$6$6	(Dividing both sides by $2$2)

 

Exploration

The following applet allows you to investigate the string and pin construction of an ellipse. You can vary the distance between the pins (foci) and change the length of the string. You can trace out the path yourself by dragging the point around (make sure the Trace button is ON), or let the animation button trace it for you.

Created with Geogebra

 

Practice questions

Question 2

question 3

 

Identify characteristics of ellipses with center $\left(0,0\right)$(0,0)

An ellipse centered at the origin has one of two equations depending on whether the longer diameter is horizontal or vertical.

If an ellipse has its major axis (longest diameter) along the $x$x-axis and its minor axis (shortest diameter) along the $y$y-axis, the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1 where:

• $a$a is half the length of the major axis
• $b$b is half the length of the minor axis

Half of the major axis is referred to as the semi-major axis, and half of the minor axis is referred to as the semi-minor axis. So we can say that $a$a is the length of the semi-major axis, and $b$b is the length of the semi-minor axis.

Graph of ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1 where $a>b$a>b.

 

If we swap the position of $a$a and $b$b, we get the equation $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2​+y2a2​=1 where $a>b$a>b with the following graph. We call these two equations the standard form of the ellipse centered at the origin.

Graph of ellipse with equation $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2​+y2a2​=1 where $a>b$a>b.

 

We call the endpoints of the major axis the vertices, and the endpoints of the minor axis the co-vertices. In the special case that the ellipse is centered at the origin, then these points represent the $x$x- and $y$y-intercepts.

Vertices and co-vertices of an ellipse.

 

If the vertices lie on the $x$x-axis, then their coordinates are $\left(-a,0\right)$(−a,0) and $\left(a,0\right)$(a,0). If they lie on the $y$y-axis then their coordinates are $\left(0,-a\right)$(0,−a) and $\left(0,a\right)$(0,a).

Similarly, if the co-vertices lie on the $x$x-axis, then they have coordinates $\left(-b,0\right)$(−b,0) and $\left(b,0\right)$(b,0). If they lie on the $y$y-axis then their coordinates are $\left(0,-b\right)$(0,−b) and $\left(0,b\right)$(0,b).

We might also be curious about identifying the foci of a given ellipse. From our definition of the ellipse, the foci represent two points on the major axis such that their sum of the distances to the ellipse is always constant.

Construction of the ellipse from two foci.

 

Recall from the definition of the ellipse, we start with the foci $\left(-c,0\right)$(−c,0) and $\left(c,0\right)$(c,0) and arrive at the equation of the ellipse where $c^2=a^2-b^2$c2=a2−b2.

Similarly, if we started with the foci $\left(0,-c\right)$(0,−c) and $\left(0,c\right)$(0,c), then we can obtain the same equation $c^2=a^2-b^2$c2=a2−b2.

Alternatively, knowing the values of $a$a and $b$b, we can find the coordinates of the two foci using the equation $c^2=a^2-b^2$c2=a2−b2.

 

Worked example

question 4

Consider the ellipse with the equation $\frac{x^2}{9^2}+\frac{y^2}{4^2}=1$x292​+y242​=1.

a) What are the coordinates of the two vertices?

Think: The vertices are the endpoints of the major axis. The major axis lies on the $x$x-axis, so the vertices will be of the form $\left(-a,0\right)$(−a,0) and $\left(a,0\right)$(a,0).

Do: Since $a=9$a=9, the vertices are $\left(-9,0\right)$(−9,0) and $\left(9,0\right)$(9,0).

b) What are the coordinates of the two co-vertices?

Think: The co-vertices are the endpoints of the minor axis. The minor axis lies on the $y$y-axis, so the vertices will be of the form $\left(0,-b\right)$(0,−b) and $\left(0,b\right)$(0,b).

Do: Since $b=4$b=4, the co-vertices are $\left(0,-4\right)$(0,−4) and $\left(0,4\right)$(0,4).

c) What are the coordinates of the foci?

Think: The distance $c$c from the center to a focus of the ellipse is related to the semi-major and semi-minor axes by the equation $c^2=a^2-b^2$c2=a2−b2. We can substitute the known values for $a$a and $b$b, and then solve for $c$c.

Do:

$c^2$c2	$=$=	$a^2-b^2$a2−b2	(Writing the equation)
$c^2$c2	$=$=	$9^2-4^2$92−42	(Substitution)
$c^2$c2	$=$=	$81-16$81−16	(Simplifying the squares)
$c^2$c2	$=$=	$65$65	(Simplifying the subtraction)
$c$c	$=$=	$\sqrt{65}$√65	(Taking the square root)

The ellipse is centered at the origin, so the coordinates of the foci are $\left(-\sqrt{65},0\right)$(−√65,0) and $\left(\sqrt{65},0\right)$(√65,0).

Reflect: How might the coordinates of the foci change if the equation of the ellipse was $\frac{x^2}{4^2}+\frac{y^2}{9^2}=1$x242​+y292​=1?

 

Practice questions

question 5

question 6

question 7

 

Ellipses with any center

To summarize characteristics above, we can start by finding a standard form for ellipses with center at $\left(0,0\right)$(0,0).

The standard form for a central ellipse depends on the orientation of the ellipse.  The equations and attributes can be summarized in the table below, given the following:

• The parameter $a$a is the length of the semi-major axis (half the length of the major axis)
• The parameter $b$b is the length of the semi-minor axis (half the length of the minor axis)
• The parameter $c$c is the distance from the center to each focus

Orientation	Horizontal Major Axis	Vertical Major Axis
Standard form	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1	$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$x2b2​+y2a2​=1
Center	$\left(0,0\right)$(0,0)	$\left(0,0\right)$(0,0)
Foci	$\left(c,0\right)$(c,0) and $\left(-c,0\right)$(−c,0)	$\left(0,c\right)$(0,c) and $\left(0,-c\right)$(0,−c)
Vertices	$\left(a,0\right)$(a,0) and $\left(-a,0\right)$(−a,0)	$\left(0,a\right)$(0,a) and $\left(0,-a\right)$(0,−a)
Covertices	$\left(0,b\right)$(0,b) and $\left(0,-b\right)$(0,−b)	$\left(b,0\right)$(b,0) and $\left(-b,0\right)$(−b,0)
Major axis	$y=0$y=0	$x=0$x=0
Minor axis	$x=0$x=0	$y=0$y=0

Notice that by this definition, it is always true that $a>b$a>b. It is also true that the parameters $a$a, $b$b, and $c$c have the relationship $c^2=a^2-b^2$c2=a2−b2.

Translated ellipses

If an ellipse is translated horizontally or vertically from the center, the parameter $a$a, $b$b, and $c$c still have the same meaning.  However, we must take into account that the center of the ellipse has moved.  Given the following definitions for $h$h and $k$k,

• The parameter $h$h denotes the translation in the horizontal direction from $0,0$0,0
• The parameter $k$k denotes the translation in the vertical direction from $0,0$0,0

The table below summarizes the standard form of an ellipse in both orientations.

Orientation	Horizontal Major Axis	Vertical Major Axis
Standard form	$\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1	$\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1
Center	$\left(h,k\right)$(h,k)	$\left(h,k\right)$(h,k)
Foci	$\left(h+c,k\right)$(h+c,k) and $\left(h-c,k\right)$(h−c,k)	$\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,k−c)
Vertices	$\left(h+a,k\right)$(h+a,k) and $\left(h-a,k\right)$(h−a,k)	$\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,k−a)
Covertices	$\left(h,k+b\right)$(h,k+b) and $\left(h,k-b\right)$(h,k−b)	$\left(h+b,k\right)$(h+b,k) and $\left(h-b,k\right)$(h−b,k)
Major axis	$y=k$y=k	$x=h$x=h
Minor axis	$x=h$x=h	$y=k$y=k

Essentially, the information is the same as the central ellipse.  But the values of $h$h and $k$k are added to the $x$x and $y$y values (respectively) for each characteristic.

 

Practice questions

question 8

question 9

question 10

 

Graphing Ellipses

It takes some skill to sketch an ellipse free hand, but with practice it does get easier.

While most modern graphing software or calculators can handle implicit forms (such as the standard form of an ellipse), it is probably a good idea to predict, with a rough sketch, what the technology should deliver. 

It can be very helpful to graph by identifying the key points such as the center, vertices, and co-vertices. We can then plot those points and sketch the ellipse. Sometimes, it may help us to draw a rectangular frame to help with the shape.

Using a rectangular frame

In standard form, the ellipse centered at $\left(0,0\right)$(0,0) has the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1. The standard form informs us that the domain is given by $\left\{x:-a\le x\le a\right\}${x:−a≤x≤a} and the range is given by $\left\{y:-b\le y\le b\right\}${y:−b≤y≤b}. We can lightly pencil in a rectangle of these dimensions to use as our guide, as shown here:

Draw in the four parts of the curve where it becomes tangent to the rectangle, then complete the sketch trying to make it as symmetric as possible as you do so. 

If the ellipse happens to be of non-central form, with the equation given by $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1  then apply the same strategy but this time with the the center of the rectangle positioned at $\left(h,k\right)$(h,k). Note that the length of the rectangle is $2a$2a and the height is $2b$2b. 

Worked example

Question 11

Sketch the graph of $\frac{\left(x-3\right)^2}{16}+\frac{\left(y-1\right)^2}{25}=1$(x−3)216​+(y−1)225​=1. 

Think: We can start by drawing a rectangle as demonstrated above, but first we need to know the coordinates of the vertices and co-vertices. It is important to note from the beginning that since the larger denominator is under the $\left(y-k\right)^2$(y−k)2 term, the major axis will be vertical.

Do: 

For this ellipse, we have the following:

• $a=\sqrt{25}$a=√25, so $a=5$a=5
• $b=\sqrt{16}$b=√16, so $b=4$b=4
• $h=3$h=3
• $k=1$k=1

From our table above, we know that the vertices are $\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,k−a). and that our covertices are $\left(h+b,k\right)$(h+b,k) and $\left(h-b,k\right)$(h−b,k).

Center	$\left(3,1\right)$(3,1)
Vertices	$\left(3,6\right)$(3,6) and $\left(3,-4\right)$(3,−4)
Covertices	$\left(7,1\right)$(7,1) and $\left(-1,1\right)$(−1,1)
Major axis	$x=3$x=3
Minor axis	$y=1$y=1

Now that we have these features, we can graph!

 

Practice questions

question 12

question 13

question 14

 

Identify domain and range of ellipses 

The domain is the set of $x$x-values of the relation and the range is the set of $y$y-values of the relation. One example of a relation is an ellipse. An ellipse that is centered at the point $\left(h,k\right)$(h,k) which has a semi-major axis of length $a$a units and semi-minor axis of length $b$b units can be described by the following equation and graph.

The graph of $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1 for $a>b$a>b.

 

The graph of the ellipse is horizontally bound by two vertical lines, one at $x=h-a$x=h−a and the other at $x=h+a$x=h+a. Because of this, the domain of the ellipse is $\left[h-a,h+a\right]$[h−a,h+a].

The ellipse bound by two vertical lines.

 

Similarly, the ellipse is bound by two vertical lines, one at $y=k-b$y=k−b and the other at $y=k+b$y=k+b and so the range of the ellipse is $\left[k-b,k+b\right]$[k−b,k+b].

The ellipse bound by two horizontal lines.

 

If we swap the length of the semi-major axis $a$a with the length of the semi-minor axis $b$b, we get the following equation and graph.

The graph of $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1 for $a>b$a>b.

 

And because we've swapped $a$a and $b$b, the domain of the ellipse becomes $\left[h-b,h+b\right]$[h−b,h+b] and the range becomes $\left[k-a,k+a\right]$[k−a,k+a].

In the special case that the ellipse is centered at the origin, that is $\left(h,k\right)=\left(0,0\right)$(h,k)=(0,0), then the domain and range of the ellipse only depend on the values of $a$a and $b$b.

Worked example

question 15

Find the domain and range of the ellipse $\frac{\left(x+3\right)^2}{25}+\frac{\left(x-5\right)^2}{16}=1$(x+3)225​+(x−5)216​=1.

Think: The equation of the ellipse is of the form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1 where $a>b$a>b. The domain of such an ellipse is given by $\left[h-a,h+a\right]$[h−a,h+a] and the range is $\left[k-b,k+b\right]$[k−b,k+b].

Do: In our case we have $h=-3$h=−3 and $a=5$a=5 so the domain of the ellipse is $\left[-3-5,-3+5\right]$[−3−5,−3+5] which simplifies to $\left[-8,2\right]$[−8,2].

We also know that $k=5$k=5 and $b=4$b=4 so the range of the ellipse is $\left[5-4,5+4\right]$[5−4,5+4] which simplifies to $\left[1,9\right]$[1,9].

Reflect: How might the domain and range differ if we swapped the semi-major and semi-minor axes?

Practice questions

question 16

question 17

 

 

Manipulate to get to standard form

Typically we are given the equation of an ellipse in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1 or $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1 where $a>b$a>b. Ideally we would like to express any equation of an ellipse in such a way, since we can immediately identify its center and the length of the semi-major and semi-minor axes:

• $\left(h,k\right)$(h,k) is the center of the ellipse.
• $a$a is the length of semi-major axis.
• $b$b is the length of semi-minor axis.

But generally speaking, the equation of an ellipse can be expressed in many ways. For instance, the following pair of equations both represent the same ellipse.

$9\left(x-2\right)^2+4\left(y-3\right)^2=36$9(x−2)2+4(y−3)2=36 or $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36

In order to express the above equations in standard form, we can manipulate both sides by multiplying or by completing the square.

 

Worked example

Question 18

Consider the equation of the ellipse $x^2-12x+36y^2=0$x2−12x+36y2=0

Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1 where $a>b$a>b.

Think: Completing the square allows us to rewrite the quadratic expressions in $x$x as a perfect square of the form $\left(x-h\right)^2$(x−h)2. Notice that the quadratic expression in $y$y is already a perfect square.

To complete the square in $x$x we look at the expression $x^2-12x$x2−12x, halve and square the coefficient of $x$x, and then add the result to both sides of the equation.

Do: Halving and squaring $-12$−12 gives a result of $36$36, so we add this result to both sides of the equation:

$x^2-12x+36+36y^2=36$x2−12x+36+36y2=36

So altogether we get the following steps:

$x^2-12x+36y^2$x2−12x+36y2	$=$=	$0$0	(Writing down the equation)
$x^2-12x+36+36y^2$x2−12x+36+36y2	$=$=	$36$36	(Completing the square for $x$x)
$\left(x-6\right)^2+36y^2$(x−6)2+36y2	$=$=	$36$36	(Factoring the perfect square)
$\frac{\left(x-6\right)^2}{36}+y^2$(x−6)236​+y2	$=$=	$1$1	(Dividing both sides by $36$36)

The equation of the ellipse is then $\frac{\left(x-6\right)^2}{36}+y^2=36$(x−6)236​+y2=36. Optionally, the standard form can be made more obvious by writing it as $\frac{\left(x-6\right)^2}{6^2}+\frac{y^2}{1^2}=36$(x−6)262​+y212​=36.

Reflect: The equation of the ellipse centered at $\left(6,0\right)$(6,0) with a vertical semi-major axis of length $6$6 units and a horizontal semi-minor axis of length $1$1 unit can be expressed by the equation $x^2-12x+36y^2=0$x2−12x+36y2=0 or the equation $\frac{\left(x-6\right)^2}{36}+y^2=1$(x−6)236​+y2=1.

 

Question 19

Consider the equation of the ellipse $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36.

Rewrite the equation in the standard form $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1 where $a>b$a>b.

Think: Completing the square allows us to rewrite the terms containing $x$x and $y$y in the form $\left(x-h\right)^2$(x−h)2 and $\left(y-k\right)^2$(y−k)2. Before we do so, it will be more convenient if we factor so that the coefficients of $x^2$x2 and $y^2$y2 are $1$1.

Do: In this case, the coefficient of $x^2$x2 is $9$9 the coefficient of $y^2$y2 is $4$4:

$9\left(x^2-4x\right)+4\left(y^2-6x\right)=-36$9(x2−4x)+4(y2−6x)=−36

Now we can complete the square in $x$x and $y$y:

$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)=-36+9\times4+4\times9$9(x2−4x+4)+4(y2−6x+9)=−36+9×4+4×9

Putting this together, we get the following steps:

$9x^2-36x+4y^2-24x$9x2−36x+4y2−24x	$=$=	$-36$−36	(Writing down the equation)
$9\left(x^2-4x\right)+4\left(y^2-6x\right)$9(x2−4x)+4(y2−6x)	$=$=	$-36$−36	(Factoring the leading coefficients of each variable)
$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9)	$=$=	$-36+9\times4+4\times9$−36+9×4+4×9	(Completing the square for $x$x and $y$y)
$9\left(x^2-4x+4\right)+4\left(y^2-6x+9\right)$9(x2−4x+4)+4(y2−6x+9)	$=$=	$36$36	(Simplifying the constant terms)
$9\left(x-2\right)^2+4\left(y-3\right)^2$9(x−2)2+4(y−3)2	$=$=	$36$36	(Rewriting the perfect squares in factored form)
$\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}$(x−2)24​+(y−3)29​	$=$=	$1$1	(Dividing both sides by $36$36)

So the equation of the ellipse can be expressed in the form $\frac{\left(x-2\right)^2}{4}+\frac{\left(y-3\right)^2}{9}=1$(x−2)24​+(y−3)29​=1.

Reflect: The center of the ellipse is $\left(h,k\right)=\left(2,3\right)$(h,k)=(2,3), the length of the vertical semi-major axis is $3$3 units and the length of the horizontal semi-minor axis is $2$2 units. From this, we can draw the equation of the ellipse as shown below.

Graph of the ellipse $9x^2-36x+4y^2-24x=-36$9x2−36x+4y2−24x=−36.

 

Practice questions

question 20

question 21

question 22

 

Applications of Ellipses

Below are two images with the important features of an ellipse:

Horizontally aligned ellipse	Vertically aligned ellipse

 

The standard form for a central ellipse depends on the orientation of the ellipse.  The equations and attributes are summarized in the table below, given the following:

• $a$a is the length of the semi-major axis.
• $b$b is the length of the semi-minor axis.
• $c$c is the distance from the center to each focus.

Notice that by this definition, it is always true that $a>b$a>b. It is also true that the parameters $a$a, $b$b, and $c$c have the relationship $c^2=a^2-b^2$c2=a2−b2.

If an ellipse is translated horizontally or vertically from the origin, the parameter $a$a, $b$b, and $c$c still have the same meaning. However, we must take into account that the center of the ellipse has moved.  Given the following definitions for $h$h and $k$k,

• $h$h denotes the translation in the horizontal direction from $\left(0,0\right)$(0,0).
• $k$k denotes the translation in the vertical direction from $\left(0,0\right)$(0,0).

An ellipse translated horizontally by $h$h and vertically by $k$k.

The table below summarizes the standard form of an ellipse in both orientations.

Orientation	Horizontal Major Axis	Vertical Major Axis
Standard form	$\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(x−h)2a2​+(y−k)2b2​=1	$\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(x−h)2b2​+(y−k)2a2​=1
Center	$\left(h,k\right)$(h,k)	$\left(h,k\right)$(h,k)
Foci	$\left(h+c,k\right)$(h+c,k) and $\left(h-c,k\right)$(h−c,k)	$\left(h,k+c\right)$(h,k+c) and $\left(h,k-c\right)$(h,k−c)
Vertices	$\left(h+a,k\right)$(h+a,k) and $\left(h-a,k\right)$(h−a,k)	$\left(h,k+a\right)$(h,k+a) and $\left(h,k-a\right)$(h,k−a)
Co-vertices	$\left(h,k+b\right)$(h,k+b) and $\left(h,k-b\right)$(h,k−b)	$\left(h+b,k\right)$(h+b,k) and $\left(h-b,k\right)$(h−b,k)
Major axis	$y=k$y=k	$x=h$x=h
Minor axis	$x=h$x=h	$y=k$y=k

Essentially, the information is the same as the central ellipse.  But the values of $h$h and $k$k are added to the $x$x and $y$y-values (respectively) for each characteristic.

 

Application to orbits

Orbits around a planetary body are described by the conic sections (circle, ellipse, parabola and hyperbola).

Elliptical orbits around the Sun. Not to scale.

The orbits of the planets around the Sun and the moon around the Earth are all elliptical (some are very close to circular though). The larger mass being orbited will be at one of the foci of the ellipse. For example, the Sun will be at one of the foci of the Earth's ellipse.

 

Worked example

question 23

At the closest point in its orbit (Perihelion) Mercury is $0.31$0.31 AU away from the Sun. At its furthest point in its orbit (Aphelion) it is $0.36$0.36 AU from the Sun. Find the equation of Mercury's orbit around the Sun. Give your values to two decimal places in AU.

Think: The equation of the ellipse will be in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$x2a2​+y2b2​=1. We can find the values of $a$a and $c$c using the geometry of the ellipse. The value of $b$b can be found given that $c^2=a^2-b^2$c2=a2−b2.

From the image above we can see that the value of $a$a will be $\frac{0.31+0.36}{2}$0.31+0.362​ and $c$c will be $\frac{0.36-0.31}{2}$0.36−0.312​.

Do: Solve for $a$a and $c$c and then sub into equation for $b$b.

$a$a	$=$=	$\frac{0.31+0.36}{2}$0.31+0.362​
	$=$=	$0.335$0.335
$c$c	$=$=	$\frac{0.36-0.31}{2}$0.36−0.312​
	$=$=	$0.025$0.025
$b$b	$=$=	$\sqrt{a^2-c^2}$√a2−c2
	$=$=	$\sqrt{0.335^2-0.025^2}$√0.3352−0.0252
	$\approx$≈	$0.334$0.334 (3 d.p.)

If we create our $xy$xy plane so the major axis is aligned with $x$x-axis and the $y$y-axis is aligned with the minor axis we can determine the equation of the orbit.

$\frac{x^2}{0.335^2}+\frac{y^2}{0.334^2}=1$x20.3352​+y20.3342​=1

 

 

Practice questions

Question 24

Question 25