12.2 The hyperbola  (Page 9/13)

$\frac{{\left(y-6\right)}^2}{36}-\frac{{\left(x+1\right)}^2}{16}=1$

$\frac{{\left(x-2\right)}^2}{49}-\frac{{\left(y+7\right)}^2}{49}=1$

$4x^2-8x-9y^2-72y+112=0$

$-9x^2-54x+9y^2-54y+81=0$

$4x^2-24x-36y^2-360y+864=0$

$-4x^2+24x+16y^2-128y+156=0$

$-4x^2+40x+25y^2-100y+100=0$

$x^2+2x-100y^2-1000y+2401=0$

$-9x^2+72x+16y^2+16y+4=0$

$4x^2+24x-25y^2+200y-464=0$

$\frac{y^2}{3^2}-\frac{x^2}{3^2}=1$

$\frac{{\left(x-3\right)}^2}{5^2}-\frac{{\left(y+4\right)}^2}{2^2}=1$

$\frac{{\left(y-3\right)}^2}{3^2}-\frac{{\left(x+5\right)}^2}{6^2}=1$

$9x^2-18x-16y^2+32y-151=0$

$16y^2+96y-4x^2+16x+112=0$

$\frac{x^2}{49}-\frac{y^2}{16}=1$

$\frac{x^2}{64}-\frac{y^2}{4}=1$

$\frac{y^2}{9}-\frac{x^2}{25}=1$

$81x^2-9y^2=1$

$\frac{{\left(y+5\right)}^2}{9}-\frac{{\left(x-4\right)}^2}{25}=1$

$\frac{{\left(x-2\right)}^2}{8}-\frac{{\left(y+3\right)}^2}{27}=1$

$\frac{{\left(y-3\right)}^2}{9}-\frac{{\left(x-3\right)}^2}{9}=1$

$-4x^2-8x+16y^2-32y-52=0$

$x^2-8x-25y^2-100y-109=0$

$-x^2+8x+4y^2-40y+88=0$

$64x^2+128x-9y^2-72y-656=0$

$16x^2+64x-4y^2-8y-4=0$

$-100x^2+1000x+y^2-10y-2575=0$

$4x^2+16x-4y^2+16y+16=0$

Vertices at $\left(3,0\right)$ and $\left(\mathrm{-3},0\right)$ and one focus at $\left(5,0\right).$

Vertices at $\left(0,6\right)$ and $\left(0,\mathrm{-6}\right)$ and one focus at $\left(0,\mathrm{-8}\right).$

Vertices at $\left(1,1\right)$ and $\left(11,1\right)$ and one focus at $\left(12,1\right).$

Center: $\left(0,0\right);$ vertex: $\left(0,\mathrm{-13}\right);$ one focus: $\left(0,\sqrt{313}\right).$

Center: $\left(4,2\right);$ vertex: $\left(9,2\right);$ one focus: $\left(4+\sqrt{26},2\right).$

Center: $\left(3,5\right);$ vertex: $\left(3,11\right);$ one focus: $\left(3,5+2\sqrt{10}\right).$

$\frac{x^2}{4}-\frac{y^2}{9}=1$

$\frac{y^2}{9}-\frac{x^2}{1}=1$

$\frac{{\left(x-2\right)}^2}{16}-\frac{{\left(y+3\right)}^2}{25}=1$

$-4x^2-16x+y^2-2y-19=0$

$4x^2-24x-y^2-4y+16=0$

The hedge will follow the asymptotes $y=x\text{ and }y=-x,$ and its closest distance to the center fountain is 5 yards.

The hedge will follow the asymptotes $y=2x\text{ and }y=\mathrm{-2}x,$ and its closest distance to the center fountain is 6 yards.

The hedge will follow the asymptotes $y=\frac{1}{2}x$ and $y=-\frac{1}{2}x,$ and its closest distance to the center fountain is 10 yards.

The hedge will follow the asymptotes $y=\frac{2}{3}x$ and $y=-\frac{2}{3}x,$ and its closest distance to the center fountain is 12 yards.

The hedge will follow the asymptotes $y=\frac{3}{4}x\text{ and }y=-\frac{3}{4}x,$ and its closest distance to the center fountain is 20 yards.

The object enters along a path approximated by the line $y=x-2$ and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line $y=-x+2.$

The object enters along a path approximated by the line $y=2x-2$ and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line $y=\mathrm{-2}x+2.$

The object enters along a path approximated by the line $y=0.5x+2$ and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line $y=\mathrm{-0.5}x-2.$

The object enters along a path approximated by the line $y=\frac{1}{3}x-1$ and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line $y=-\frac{1}{3}x+1.$

The object It enters along a path approximated by the line $y=3x-9$ and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line $y=\mathrm{-3}x+9.$