Circles and Ellipses

Lesson

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

While most modern graphing packages 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 package should deliver. Just relying on the package might be problematic - after all they are really only plotting machines.

By understanding a few construction concepts, we can make sense of, and can be come more confident in what a graphing screen shows us.

Here are a few strategies:

In standard form the central ellipse has the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$`x`2`a`2+`y`2`b`2=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`)2`a`2+(`y`−`k`)2`b`2=1 then apply the same strategy but this time with the the centre of the rectangle positioned at $\left(h,k\right)$(`h`,`k`) (see example in strategy 3). Note that the length of the rectangle is $2a$2`a` and the height is $2b$2`b`.

Note that for the central ellipse, with centre at the origin, the foci in general are located at $x=\pm ae$`x`=±`a``e`. Since $b^2=a^2\left(1-e^2\right)$`b`2=`a`2(1−`e`2) then $e=\frac{\sqrt{a^2-b^2}}{a}$`e`=√`a`2−`b`2`a` and hence $ae=\sqrt{a^2-b^2}$`a``e`=√`a`2−`b`2. In other words the distance that each focus is away from the centre of an ellipse is given by $\sqrt{a^2-b^2}$√`a`2−`b`2.

A nice little trick is to utilize Pythagoras' Theorem in order to geometrically find the correct position of the foci.

Take a strip of paper and mark off, from the left edge of the strip, the length of the semi-major axis $OA$`O``A`. Then lay the strip diagonally with the left edge of the strip directly over $B$`B` and rotate the strip until the length $OA$`O``A` coincides with the major axis, as shown in this diagram.

The length $OS$`O``S` from the diagram is, by Pythagoras' Theorem, $\sqrt{a^2-b^2}$√`a`2−`b`2. Shift the paper strip across to locate the other focus.

We are often presented with an ellipse that is not in standard form. For example, an ellipse in non-standard form may be given as $9x^2+25y^2=225$9`x`2+25`y`2=225.

If we divide the expression by the number on the right hand side, $225$225, we immediately reveal the standard form $\frac{x^2}{25}+\frac{y^2}{9}=1$`x`225+`y`29=1. We can then use the rectangle and paper strip idea to sketch the ellipse and locate the foci.

We can also find the $x$`x` and $y$`y` intercepts in the usual way.

Putting $y=0$`y`=0 we have $9x^2=225$9`x`2=225 or that $x^2=25$`x`2=25 and thus $x=\pm5$`x`=±5. Putting $x=0$`x`=0 we have $25y^2=225$25`y`2=225 or that $y^2=9$`y`2=9 and thus $y=\pm3$`y`=±3. Hence the $x$`x` and $y$`y` intercepts are given as $\left(\pm5,0\right)$(±5,0) and $\left(0,\pm3\right)$(0,±3) respectively.

The equation $25x^2+36y^2+150x-72y=639$25`x`2+36`y`2+150`x`−72`y`=639 represents a non-centric ellipse. We can show this by completing the squares on both x and y as follows:

$25x^2+36y^2+150x-72y$25x2+36y2+150x−72y |
$=$= | $639$639 |

$25\left(x^2+6x\right)+36\left(y^2-2y\right)$25(x2+6x)+36(y2−2y) |
$=$= | $639$639 |

$25\left(x^2+6x+9\right)+36\left(y^2-2y+1\right)$25(x2+6x+9)+36(y2−2y+1) |
$=$= | $639+225+36$639+225+36 |

$25\left(x+3\right)^2+36\left(y-1\right)^2$25(x+3)2+36(y−1)2 |
$=$= | $900$900 |

$\frac{25\left(x+3\right)^2}{900}+\frac{36\left(y-1\right)^2}{900}$25(x+3)2900+36(y−1)2900 |
$=$= | $1$1 |

$\frac{\left(x+3\right)^2}{36}+\frac{\left(y-1\right)^2}{25}$(x+3)236+(y−1)225 |
$=$= | $1$1 |

With $a=6$`a`=6 and $b=5$`b`=5, our rectangle has dimensions $12$12 by $10$10. We pencil in the rectangle to have a centre of $\left(-3,1\right)$(−3,1) and fill in the sketch as follows:

We could use the paper strip idea to locate the position of the foci. If you need to , you can determine their exact coordinates.

For example, with this ellipse, using $b^2=a^2\left(1-e^2\right)$`b`2=`a`2(1−`e`2), we have that $25=36\left(1-e^2\right)$25=36(1−`e`2) and this can be solve to $e=\frac{\sqrt{11}}{6}$`e`=√116. This means that the quantity $ae=\sqrt{11}$`a``e`=√11 and the foci are positioned at $\left(-3\pm\sqrt{11},1\right)$(−3±√11,1).

Sketch the graph of the equation $\frac{x^2}{25}+\frac{y^2}{16}=1$`x`225+`y`216=1

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Consider the ellipse with equation $x^2+25y^2=25$`x`2+25`y`2=25.

Write the equation in standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(

`x`−`h`)2`a`2+(`y`−`k`)2`b`2=1 where $a>b$`a`>`b`.Draw the graph of the ellipse.

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Consider the ellipse with equation $4x^2+25y^2-16x+150y+141=0$4`x`2+25`y`2−16`x`+150`y`+141=0.

Rewrite the equation in standard form $\frac{\left(x-h\right)^2}{a^2}+\frac{\left(y-k\right)^2}{b^2}=1$(

`x`−`h`)2`a`2+(`y`−`k`)2`b`2=1 where $a>b$`a`>`b`.Draw the graph of the ellipse.

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Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems