Circles and Ellipses

Lesson

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`)2`a`2+(`y`−`k`)2`b`2=1 or $\frac{\left(x-h\right)^2}{b^2}+\frac{\left(y-k\right)^2}{a^2}=1$(`x`−`h`)2`b`2+(`y`−`k`)2`a`2=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 centre and the length of the semi-major and semi-minor axes:

- $\left(h,k\right)$(
`h`,`k`) is the centre 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$9`x`2−36`x`+4`y`2−24`x`=−36

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

Consider the equation of the ellipse $x^2-12x+36y^2=0$`x`2−12`x`+36`y`2=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`)2`a`2+(`y`−`k`)2`b`2=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 $x^2-12x$`x`2−12`x`, **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:

$x^2-12x+36+36y^2=36$`x`2−12`x`+36+36`y`2=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 | (Factorising 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+`y`2=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+`y`212=36.

**Reflect:** The equation of the ellipse centred 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 $x^2-12x+36y^2=0$`x`2−12`x`+36`y`2=0 or the equation $\frac{\left(x-6\right)^2}{36}+y^2=1$(`x`−6)236+`y`2=1.

Consider the equation of the ellipse $9x^2-36x+4y^2-24x=-36$9`x`2−36`x`+4`y`2−24`x`=−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`)2`b`2+(`y`−`k`)2`a`2=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 factorise so that the coefficients of $x^2$`x`2 and $y^2$`y`2 are $1$1.

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

$9\left(x^2-4x\right)+4\left(y^2-6x\right)=-36$9(`x`2−4`x`)+4(`y`2−6`x`)=−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(`x`2−4`x`+4)+4(`y`2−6`x`+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 | (Factorising 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 factorised 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 centre 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. |

Consider the ellipse with equation $\left(x+3\right)^2+4\left(y-1\right)^2=16$(`x`+3)2+4(`y`−1)2=16.

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.

Loading Graph...

The graph of the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$`x`29+`y`24=1 is given below.

Loading Graph...

Write $4x^2+9\left(y-2\right)^2=36$4

`x`2+9(`y`−2)2=36 in the 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`.What transformation of the ellipse drawn gives the ellipse $4x^2+9\left(y-2\right)^2=36$4

`x`2+9(`y`−2)2=36?Translated to the right by $2$2 units

ATranslated to the left by $2$2 units

BTranslated up by $2$2 units

CTranslated down by $2$2 units

DTranslated to the right by $2$2 units

ATranslated to the left by $2$2 units

BTranslated up by $2$2 units

CTranslated down by $2$2 units

D

Consider the ellipse with equation $4x^2+24x+5y^2+20y+36=0$4`x`2+24`x`+5`y`2+20`y`+36=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`)2`a`2+(`y`−`k`)2`b`2=1 where $a>b$`a`>`b`.What are the coordinates of the centre of the ellipse?

What are the coordinates of the vertices?

Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems