Hong Kong

Stage 4 - Stage 5

Lesson

The inequality $\frac{x^2}{a^2}-\frac{y^2}{b^2}<1$`x`2`a`2−`y`2`b`2<1 includes the open region within the two hyperbolic arcs given by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$`x`2`a`2−`y`2`b`2=1.

A nice way to remember this is to think about the solution to the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$`x`2`a`2−`y`2`b`2=0. Only when $y=\pm\frac{b}{a}x$`y`=±`b``a``x` will this equation be true, and these are the oblique asymptotes of the hyperbola. Thus, since $0<1$0<1, the asymptotes are included in the region $\frac{x^2}{a^2}-\frac{y^2}{b^2}<1$`x`2`a`2−`y`2`b`2<1. That is to say the points satisfying the inequality $\frac{x^2}{a^2}-\frac{y^2}{b^2}<1$`x`2`a`2−`y`2`b`2<1 will form the open region *within* the arcs.

Here is the graph of the inequality relation $\frac{x^2}{9}-\frac{y^2}{16}<1$`x`29−`y`216<1. Note we are using the convention of shading out (the pink area) the region *not satisfying* the inequality. Also note the asymptotes $y=\pm\frac{4}{3}x$`y`=±43`x` are completely inside the region of interest:

A more challenging example might involve determining the region satisfying more than one inequality. Consider the following:

Determine the closed region defined by the intersection of the three inequalities:

- $y\le1.5x^2$
`y`≤1.5`x`2 - $y\ge-1.5x^2$
`y`≥−1.5`x`2 - $x^2-\frac{y^2}{9}\le1$
`x`2−`y`29≤1

To begin with, we need to determine where each of the two boundary parabolae intersect the hyperbola. For the given question, we know that the region we seek is closed, and so intersections are expected.

To do this, we solve simultaneously the parabola $y=1.5x^2$`y`=1.5`x`2 with the hyperbola $x^2-\frac{y^2}{9}=1$`x`2−`y`29=1.

Since $y=1.5x^2$`y`=1.5`x`2, we have that $x^2=\frac{2}{3}y$`x`2=23`y`, and so upon substitution, we need to solve for $y$`y` in the equation $\left(\frac{2}{3}y\right)-\frac{y^2}{9}=1$(23`y`)−`y`29=1. Thus:

$\left(\frac{2}{3}y\right)-\frac{y^2}{9}$(23y)−y29 |
$=$= | $1$1 |

$6y-y^2$6y−y2 |
$=$= | $9$9 |

$y^2-6y+9$y2−6y+9 |
$=$= | $0$0 |

$\left(y-3\right)^2$(y−3)2 |
$=$= | $0$0 |

$\therefore$∴ $y$y |
$=$= | $3$3 |

There is only one $y$`y` value, meaning that the parabola *must be tangent* to the hyperbola (see diagram below).

Since $x^2=\frac{2}{3}y$`x`2=23`y`, we have that $x^2=2$`x`2=2 and $x=\pm\sqrt{2}$`x`=±√2. That is to say the two points of tangency are located at $\left(\pm\sqrt{2},3\right)$(±√2,3).

Similarly, because of symmetry, the two points of tangency with the inverted parabola $y=-1.5x^2$`y`=−1.5`x`2 and the hyperbola must be $\left(\pm\sqrt{2},-3\right)$(±√2,−3).

The interesting intersection of the three inequalities is defined as the region between the arcs of the hyperbola and above and below the inverted parabola and upward opening parabola respectively.