NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Manipulation of Hyperbolic Equations

Lesson

Typically we are given the equation of a hyperbola 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 $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$(`y`−`k`)2`a`2−(`x`−`h`)2`b`2=1. Ideally, we would like to express any equation of a hyperbola in this form since we can immediately identify its centre, orientation, and other features.

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

$4\left(x-2\right)^2-25\left(y-3\right)^2=100$4(`x`−2)2−25(`y`−3)2=100 and $x^2-4x-y^2+6y=105$`x`2−4`x`−`y`2+6`y`=105

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

Consider the equation of the hyperbola $x^2-4x-y^2+6y=105$`x`2−4`x`−`y`2+6`y`=105.

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.

**Think:** Completing the square allows us to rewrite the quadratic expressions in $x$`x` and $y$`y` to be perfect squares of 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 the coefficient of $y^2$`y`2 is $1$1.

$x^2-4x-\left(y^2-6y\right)=105$`x`2−4`x`−(`y`2−6`y`)=105

Then to complete the square in $x$`x` we look at the expression $x^2-4x$`x`2−4`x`, **halve** and **square** the coefficient of $x$`x`, and then add the result to both sides of the equation.

**Do:** Halving and squaring $-4$−4 gives a result of $4$4, so we add this result to both sides of the equation:

$x^2-4x+4-\left(y^2-6y\right)=105+4$`x`2−4`x`+4−(`y`2−6`y`)=105+4

Similarly, to complete the square in $y$`y` we add $9$9 inside the brackets:

$x^2-4x+4-\left(y^2-6y+9\right)=105+4-9$`x`2−4`x`+4−(`y`2−6`y`+9)=105+4−9

Because of the negative sign outside of the brackets, we will subtract $9$9 from the other side to balance the equation.

Putting this together, we get the following steps:

$x^2-4x-y^2+6y$x2−4x−y2+6y |
$=$= | $105$105 | (Writing down the equation) |

$x^2-4x-\left(y^2-6y\right)$x2−4x−(y2−6y) |
$=$= | $105$105 | (Factorising the terms with $y$y) |

$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9) |
$=$= | $105+4-9$105+4−9 | (Completing the square for $x$x and $y$y) |

$x^2-4x+4-\left(y^2-6y+9\right)$x2−4x+4−(y2−6y+9) |
$=$= | $100$100 | (Simplifying the constant terms) |

$\left(x-2\right)^2-\left(y-3\right)^2$(x−2)2−(y−3)2 |
$=$= | $100$100 | (Factorising the perfect squares) |

$\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}$(x−2)2100−(y−3)2100 |
$=$= | $1$1 | (Dividing both sides by $100$100) |

The equation of the hyperbola is then $\frac{\left(x-2\right)^2}{100}-\frac{\left(y-3\right)^2}{100}=1$(`x`−2)2100−(`y`−3)2100=1 which is in the 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.

**Reflect:** The hyperbola with equation $x^2-4x-y^2+6y=105$`x`2−4`x`−`y`2+6`y`=105 has a centre of $\left(2,3\right)$(2,3) with vertices $\left(-8,3\right)$(−8,3) and $\left(12,3\right)$(12,3). What are the coordinates of the foci?

Consider the equation of the circle $3y^2+12y-x^2-6x=3$3`y`2+12`y`−`x`2−6`x`=3.

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

**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 the coefficients of $x^2$`x`2 and $y^2$`y`2 were $1$1.

**Do:** In this case, the coefficient of $x^2$`x`2 is $-1$−1 and the coefficient of $y^2$`y`2 is $3$3, so we factorise out these coefficients:

$3\left(y^2+4y\right)-\left(x^2+6x\right)=3$3(`y`2+4`y`)−(`x`2+6`x`)=3

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

$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)=3+3\times4-9$3(`y`2+4`y`+4)−(`x`2+6`x`+9)=3+3×4−9

Notice that we've added by $3\times4$3×4 and subtracted by $9$9 on the other side to balance the equation. This is needed because the terms that we've added on the left hand side are being multiplied by $3$3 and $-1$−1 respectively.

Putting this together, we get the following steps:

$3y^2+12y-x^2-6x$3y2+12y−x2−6x |
$=$= | $3$3 | (Writing down the equation) |

$3\left(y^2+4y\right)-\left(x^2+6x\right)$3(y2+4y)−(x2+6x) |
$=$= | $3$3 | (Factorising the terms with $x$x and $y$y) |

$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9) |
$=$= | $3+3\times4-9$3+3×4−9 | (Completing the square for $x$x and $y$y) |

$3\left(y^2+4y+4\right)-\left(x^2+6x+9\right)$3(y2+4y+4)−(x2+6x+9) |
$=$= | $6$6 | (Simplifying the constant terms) |

$3\left(y+2\right)^2-\left(x+3\right)^2$3(y+2)2−(x+3)2 |
$=$= | $6$6 | (Factorising the perfect squares) |

$\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}$(y+2)22−(x+3)26 |
$=$= | $1$1 | (Dividing both sides by $6$6) |

So the equation of the hyperbola can be expressed in the form $\frac{\left(y+2\right)^2}{2}-\frac{\left(x+3\right)^2}{6}=1$(`y`+2)22−(`x`+3)26=1 which is in the form $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(h-k\right)^2}{b^2}=1$(`y`−`k`)2`a`2−(`h`−`k`)2`b`2=1.

**Reflect:** The centre of the hyperbola is $3y^2+12y-x^2-6x=3$3`y`2+12`y`−`x`2−6`x`=3 is $\left(-3,-2\right)$(−3,−2) and has vertices $\left(-3,-2+\sqrt{2}\right)$(−3,−2+√2) and $\left(-3,-2-\sqrt{2}\right)$(−3,−2−√2). What are the coordinates of the foci?

Consider the hyperbola with equation $4y^2-9x^2=36$4`y`2−9`x`2=36.

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

`y`−`k`)2`a`2−(`x`−`h`)2`b`2=1.What are the coordinates of the centre?

What are the coordinates of the vertices?

What are the coordinates of the foci?

What are the equations of the asymptotes?

Consider the hyperbola with equation $y^2+2y-7x^2-6=0$`y`2+2`y`−7`x`2−6=0.

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

`y`−`k`)2`a`2−(`x`−`h`)2`b`2=1.What are the coordinates of the centre?

What are the coordinates of the vertices?

Consider the hyperbola with equation $2x^2+8x-5y^2-10y=7$2`x`2+8`x`−5`y`2−10`y`=7.

Write 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.What are the coordinates of the centre?

What are the coordinates of the vertices?

Apply the geometry of conic sections

Apply the geometry of conic sections in solving problems