NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Domain and Range of Ellipses
Lesson

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 centred 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=ha and the other at $x=h+a$x=h+a. Because of this, the domain of the ellipse is $\left[h-a,h+a\right]$[ha,h+a].

 The ellipse bound by two vertical lines.

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

Remember!

For an ellipse of the form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(xh)2a2+(yk)2b2=1 where $a>b$a>b:

The domain is equal to $\left[h-a,h+a\right]$[ha,h+a] and the range is equal to $\left[k-b,k+b\right]$[kb,k+b].

For an ellipse of the form $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(xh)2b2+(yk)2a2=1 where $a>b$a>b:

The domain is equal to $\left[h-b,h+b\right]$[hb,h+b] and the range is equal to $\left[k-a,k+a\right]$[ka,k+a].

In the special case that the ellipse is centred 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

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+(x5)216=1.

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

Do: In our case we have $h=-3$h=3 and $a=5$a=5 so the domain of the ellipse is $\left[-3-5,-3+5\right]$[35,3+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 $\left[5-4,5+4\right]$[54,5+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 1

Consider the graph of the ellipse drawn below.

1. What is the domain of the ellipse?

Write the answer as an interval.

2. What is the range of the ellipse?

Write the answer as an interval.

##### question 2

Consider the equation of the ellipse $\frac{\left(x+6\right)^2}{6^2}+\frac{\left(y+3\right)^2}{4^2}=1$(x+6)262+(y+3)242=1.

1. Draw the graph of the ellipse.

2. What is the domain of the ellipse?

Write the answer as an interval.

3. What is the range of the ellipse?

Write the answer as an interval.

##### question 3

Consider the equation of the ellipse $\frac{\left(x+3\right)^2}{9^2}+\frac{\left(y+2\right)^2}{18^2}=1$(x+3)292+(y+2)2182=1.

1. What is the domain of the ellipse?

Write the answer as an interval.

2. What is the range of the ellipse?

Write the answer as an interval.

### Outcomes

#### M8-1

Apply the geometry of conic sections

#### 91573

Apply the geometry of conic sections in solving problems