$y^2=-6x$

$y^2=12x$

$x^2=7y$

$x^2=-3y$

${(x-2)}^2=-(y+3)$

${(y+1)}^2=5(x+2)$

$y^2=-4y+2x+1$

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

Vertex: (0, 0) focus: $(-3,0)$

Vertex: (0, 0) focus: (0, 4)

Focus: (0, 4) directrix: $y=-4$

Focus: $(-3,0);$ directrix: $x=3$

${\displaystyle \frac{x^2}{25}}+{\displaystyle \frac{y^2}{4}}=1$

${\displaystyle \frac{x^2}{9}}+{\displaystyle \frac{y^2}{36}}=1$

$4x^2+y^2=4$

$16x^2+y^2=64$

$16{(x+1)}^2+9{(y+4)}^2=144$

$4{(x-1)}^2+3{(y+2)}^2=12$

$x^2+9y^2+2x-18y+1=0$

$4x^2+y^2+8x-10y+13=0$

Vertices: $(\pm 4,0);$ endpoints of minor axis: $(0,\pm 2)$

Vertices: $(0,\pm 6)$; endpoints of minor axis: $(\pm 2,0)$

Length of major axis: 20; foci: $(\pm 5,0)$

Length of minor axis: 16; foci: $(0,\pm 6)$

${\displaystyle \frac{y^2}{16}}-{\displaystyle \frac{x^2}{4}}=1$

${\displaystyle \frac{x^2}{16}}-{\displaystyle \frac{y^2}{9}}=1$

$8x^2-y^2=8$

$4y^2-4x^2=1$

${\displaystyle \frac{{(x+2)}^2}{9}}-{\displaystyle \frac{{(y-3)}^2}{4}}=1$

${\displaystyle \frac{{(y+1)}^2}{6}}-{\displaystyle \frac{{(x-2)}^2}{8}}=1$

$4y^2-x^2+40y-4x+60=0$

$4x^2-9y^2+16x-54y-29=0$

Vertices: $(\pm 1,0);$ foci: $(\pm 2,0)$

Vertices: $(0,\pm 2);$ foci: $(0,\pm 4)$

Vertices: $(\pm 2,0);$ asymptotes: $y=\pm 3x$

Vertices: $(0,\pm 3);$ asymptotes: $y=\pm x$

$5x^2-4y^2=20$

$x^2-x+y=1$

$3x^2+4y^2+8y-12x-6=0$

$2y-x^2=0$

$x^2+y^2+2x-3=0$

$2x+y^2=0$

$y^2=x^2+3$

$x^2=10-3y^2$

$3x^2+3y^2-6x+12y+5=0$

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

$9x^2+8y^2=36$

$2y^2+4y=3x^2-6x+9$

Find an equation of the hyperbola whose foci are the vertices of the ellipse $4x^2+9y^2=36$ and whose vertices are the foci of this ellipse.

Find an equation of the ellipse whose foci are the vertices of the hyperbola $9x^2-16y^2=144$ and whose vertices are the foci of this hyperbola.

$x^2-4y^2=36$ and $x-2y-20=0$

$y^2-8x^2=5$ and $y-2x^2=0$

$3x^2-7y^2=5$ and $9y^2-2x^2=1$

$x^2-y^2=1$ and $x^2+y^2=7$

Parabolic curve. Water flowing from the end of a horizontal pipe 20 feet above the ground describes a parabolic curve whose vertex is at the end of the pipe. If at a point 6 feet below the line of the pipe the flow of water has curved outward 8 feet beyond a vertical line through the end of the pipe, how far beyond this vertical line will the water strike the ground?

Parabolic arch. A parabolic arch has a height of 20 meters and a width of 36 meters at the base. Assuming that the vertex of the parabola is at the top of the arch, find the height of the arch at a distance of 9 meters from the center of the base.

Football. A football is 12 inches long, and a cross section containing a seam is an ellipse with a minor axis of length 7 inches. Suppose every cross section of the ball formed by a plane perpendicular to the major axis of the ellipse is a circle. Find the circumference of such a circular cross section located 2 inches from an end of the ball.

Production cost. A company assembles computers at two locations A and B that are 1000 miles apart. The per unit cost of production at location A is $20 less than at location B. Assume that the route of delivery of the computers is along a straight line and that the delivery cost is $25¢$ per unit per mile. Find the equation of the curve at any point the computers can be supplied from either location at the same cost.

[Hint: Take A at $(-500,0)$ and B at (500, 0).]