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Stage 4 - Stage 5

Lesson

Given enough information about a conic, it is possible to find its eccentricity.

For the circle, the eccentricity is $0$0 and for the parabola the eccentricity is $1$1. The other two conics - the ellipse and the hyperbola have eccentricities within the ranges $0`e`<1 and $e>1$`e`>1 respectively. So if you happen to encounter a conic with an eccentricity of $1000$1000, then its sure to be an hyperbola!

These two ellipses have the same eccentricity:

They are both stretched the same amount but the *direction* of stretch is different. They both have the same general equation form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$`x`2`a`2+`y`2`b`2=1, but the difference is that in the top ellipse the size of $a$`a` (the length of the semi-major axis) is larger than the size of $b$`b` (the length of the semi minor axis). In the bottom ellipse it is the other way around.

The formula for the eccentricity of the top ellipse, as we have met before, is given by $e=\frac{\sqrt{a^2-b^2}}{a}$`e`=√`a`2−`b`2`a` and the formula for the bottom ellipse is $e=\frac{\sqrt{b^2-a^2}}{b}$`e`=√`b`2−`a`2`b`. While they look different formulae, they will in fact deliver the same result.

This is because both formulae really state exactly the same relationship. If we call the length of the semi-major axis $sM$`s``M` and the length of the semi-minor axis $sm$`s``m`, then for both ellipses:

$e=\frac{\sqrt{\left(sM\right)^2-\left(sm\right)^2}}{sM}$`e`=√(`s``M`)2−(`s``m`)2`s``M`

Clearly we would be in all sorts of trouble if, under the square root, a larger square was being subtracted from a smaller square.

The standard formula for the hyperbola is given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=1, and its eccentricity is given by $e=\frac{\sqrt{a^2+b^2}}{a}$`e`=√`a`2+`b`2`a`. Note the similarity to the ellipse, except the signs are flipped around in both formulae.

The other big difference is that it doesn't matter about the relative sizes of $a$`a` and $b$`b`, as it did with the ellipse - the same formula applies. The foci will always lie along the *horizontal axis*, no matter what the values of $a$`a` or $b$`b` are.

However, a new hyperbola - the conjugate hyperbola - appears if we swap the position of the two squares in the standard formula. Specifically, the conjugate hyperbola has the equation $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$`y`2`b`2−`x`2`a`2=1. Only in this case does the eccentricity formula change to $e=\frac{\sqrt{a^2+b^2}}{b}$`e`=√`a`2+`b`2`b`.

For the conjugate hyperbola, the foci similarly will always lie along the *vertical axis*, no matter what the value of $a$`a` and $b$`b` are.

The two possibilities are depicted here:

Note that to find $e$`e`, we either need $a$`a` and $b$`b` or we need enough information to find $a$`a` and $b$`b`.

The information may be given algebraically, or graphically, or simply stated in a sentence. We will demonstrate with a few examples:

$9x^2+100y^2=900$9`x`2+100`y`2=900

This is an ellipse, and so:

$9x^2+100y^2$9x2+100y2 |
$=$= | $900$900 |

$\frac{x^2}{100}+\frac{y^2}{9}$x2100+y29 |
$=$= | $1$1 |

This means $a=10$`a`=10 and $b=3$`b`=3. The ellipse is elongated horizontally, with $e=\frac{\sqrt{100-9}}{10}=\frac{\sqrt{91}}{10}$`e`=√100−910=√9110.

$4y^2-16x^2=256$4`y`2−16`x`2=256

This is the equation of a conjugate hyperbola, and by rearranging it into standard form we have:

$4y^2-16x^2$4y2−16x2 |
$=$= | $256$256 |

$\frac{y^2}{64}-\frac{x^2}{16}$y264−x216 |
$=$= | $1$1 |

Therefore $a=4$`a`=4 and $b=8$`b`=8 (check this). Hence the eccentricity is given by $e=\frac{\sqrt{4^2+8^2}}{8}$`e`=√42+828 simplifying to $e=\frac{\sqrt{5}}{2}$`e`=√52.

A certain central conic of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$`x`2`a`2+`y`2`b`2=1 passes through the point $\left(4,6\right)$(4,6) and has a semi-major axis length of $10$10. Show that, without a diagram, there are two possibilities for the eccentricity.

Based on the information given, and without a diagram, we are not sure whether $a$`a` or $b$`b` represents the length of the semi-major axis.

So there are in fact two possibilities for the equation, both ellipses, given as $\frac{x^2}{10^2}+\frac{y^2}{b^2}=1$`x`2102+`y`2`b`2=1 (elongated horizontally) or $\frac{x^2}{a^2}+\frac{y^2}{10^2}=1$`x`2`a`2+`y`2102=1 (elongated vertically).

Taking the first option, $\frac{x^2}{100}+\frac{y^2}{b^2}=1$`x`2100+`y`2`b`2=1, and knowing that $\left(4,6\right)$(4,6) is a point on the conic, we have:

$\frac{4^2}{100}+\frac{6^2}{b^2}$42100+62b2 |
$=$= | $1$1 |

$\frac{6^2}{b^2}$62b2 |
$=$= | $1-\frac{4^2}{100}$1−42100 |

$\frac{6^2}{b^2}$62b2 |
$=$= | $\frac{84}{100}$84100 |

$\therefore$∴ $b^2$b2 |
$=$= | $\frac{300}{7}$3007 |

This means that the ellipse could have the equation $\frac{x^2}{100}+\frac{7y^2}{300}=1$`x`2100+7`y`2300=1, with $a=10$`a`=10 and $b=\frac{10\sqrt{3}}{\sqrt{7}}$`b`=10√3√7. The eccentricity would be given by $e=\frac{\sqrt{100-\frac{300}{7}}}{10}$`e`=√100−300710 which when simplified becomes $e=\frac{2}{\sqrt{7}}$`e`=2√7.

On the other hand, the ellipse could have the equation $\frac{x^2}{a^2}+\frac{y^2}{10^2}=1$`x`2`a`2+`y`2102=1, and so proceeding in a similar manner with the point $\left(4,6\right)$(4,6), we have:

$\frac{16}{a^2}+\frac{36}{100}$16a2+36100 |
$=$= | $1$1 |

$\frac{16}{a^2}$16a2 |
$=$= | $1-\frac{36}{100}$1−36100 |

$\frac{16}{a^2}$16a2 |
$=$= | $\frac{64}{100}$64100 |

$\therefore$∴ $a^2$a2 |
$=$= | $25$25 |

The ellipse would then have the equation $\frac{x^2}{25}+\frac{y^2}{100}=1$`x`225+`y`2100=1, with $a=5$`a`=5 and $b=10$`b`=10. The eccentricity would then be given by $e=\frac{\sqrt{100-25}}{10}=\frac{\sqrt{75}}{10}$`e`=√100−2510=√7510, simplifying to $e=\frac{\sqrt{3}}{2}$`e`=√32.

Here are the two possibilities:

Consider the hyperbola with the equation $16x^2-9y^2=144$16`x`2−9`y`2=144.

Rewrite the equation in the standard form for a hyperbola.

Hence find the eccentricity $e$

`e`of the hyperbola.

The point $P\left(\frac{17}{2},-\frac{45}{8}\right)$`P`(172,−458) lies on a conic section with a focus at $\left(5,0\right)$(5,0) and directrix at $x=\frac{16}{5}$`x`=165.

Find the eccentricity of the conic section.

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Consider the hyperbola with the equation $4x^2-y^2=16$4`x`2−`y`2=16.

Rewrite the equation in the standard form for a hyperbola.

Hence find the eccentricity $e$

`e`of the hyperbola.