Lesson

A line or curve can be thought of as the locus (or path) of all points that satisfy the line or curve's equation. A point with coordinates $\left(x_1,y_1\right)$(`x`1,`y`1) is said to satisfy the equation if, when $x_1$`x`1 and $y_1$`y`1 are substituted into the line or curve's equation, it makes that equation true.

For example, we know that $\left(2,0\right)$(2,0) lies on the curve given by $4x^2-y^2=16$4`x`2−`y`2=16, simply because $4\times\left(2\right)^2-\left(0\right)^2=16$4×(2)2−(0)2=16.

Since every point on a curve satisfies its equation, then we can use this fact to determine the unknown coefficients of a curve's equation.

Suppose we know that $\left(2,-3\right)$(2,−3) is a point on the curve $y=\frac{k}{x}$`y`=`k``x`. Then we can argue as follows:

$y$y |
$=$= | $\frac{k}{x}$kx |

$-3$−3 | $=$= | $\frac{k}{2}$k2 |

$k$k |
$=$= | $-6$−6 |

Thus the equation of the curve is $y=-\frac{6}{x}$`y`=−6`x`, and all points on the curve will satisfy it.

The curve should be recognised as a hyperbola, with a dilation factor of $-6$−6 and possessing mutually perpendicular asymptotes of $x=0$`x`=0 and $y=0$`y`=0. The arcs of the hyperbola lie in the upper left and lower right quadrants formed by these asymptotes.

With the equation uniquely defined, we can now easily find other points that satisfy it. For example $\left(1,-6\right)$(1,−6) and $\left(3,-2\right)$(3,−2) are two more points on the curve.

Suppose we know that the points $\left(9,9\right)$(9,9) and $\left(1,5\right)$(1,5) are points on the hyperbola given by $y=\frac{a}{x-3}+k$`y`=`a``x`−3+`k`.

In this instance there are two points given and there are two unknowns in the given equation. By using the above substitution strategy twice, we should be able to ascertain the values of the unknown constants.

From the first point, we substitute and simply as follows:

$y$y |
$=$= | $\frac{a}{x-3}+k$ax−3+k |

$9$9 | $=$= | $\frac{a}{9-3}+k$a9−3+k |

$9-k$9−k |
$=$= | $\frac{a}{6}$a6 |

$54-6k$54−6k |
$=$= | $a$a |

$a+6k$a+6k |
$=$= | $54$54 |

In the same way we substitute the other point to find a second equation:

$y$y |
$=$= | $\frac{a}{x-3}+k$ax−3+k |

$5$5 | $=$= | $\frac{a}{1-3}+k$a1−3+k |

$5-k$5−k |
$=$= | $\frac{a}{-2}$a−2 |

$2k-10$2k−10 |
$=$= | $a$a |

$a-2k$a−2k |
$=$= | $-10$−10 |

Thus we have two simultaneous equations that we can solve as follows:

$a+6k$a+6k |
$=$= | $54$54 |

$a-2k$a−2k |
$=$= | $-10$−10 |

$\therefore8k$∴8k |
$=$= | $64$64 |

$k$k |
$=$= | $8$8 |

$\therefore a+48$∴a+48 |
$=$= | $54$54 |

$a$a |
$=$= | $6$6 |

Thus the hyperbola has the equation $y=\frac{6}{x-3}+8$`y`=6`x`−3+8. The curve is a hyperbola, with a dilation factor of $6$6 and possessing mutually perpendicular asymptotes of $x=3$`x`=3 and $y=8$`y`=8. The arcs of the hyperbola lie in the lower left and upper right quadrants formed by these asymptotes.

The point $P$`P`$\left(1,4\right)$(1,4) lies on the hyperbola with equation $y=\frac{k}{x}$`y`=`k``x`. Find the value of $k$`k`, and hence find the equation of the hyperbola shown.

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The point $P$`P`$\left(-1,-9\right)$(−1,−9) lies on the hyperbola with equation $y=\frac{k}{x}$`y`=`k``x`. Find the value of $k$`k`, and hence find the equation of the hyperbola shown.

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The graph of $y=\frac{3}{x}$`y`=3`x` is rotated $90^\circ$90° clockwise about the origin.

What will be the equation of the new graph?

A point on the hyperbola $y=\frac{3}{x}$

`y`=3`x` is $\left(1,3\right)$(1,3). What will be the new point after the rotation?$($( $\editable{}$, $\editable{}$ $)$)

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

Apply graphical methods in solving problems