In Exercises parabolasketchfirst - parabolasketchlast, graph the given equations in the \(xy\)-plane. Find the vertex, focus and directrix. Include the endpoints of the latus rectum in your sketch.
\((x - 3)^{2} = -16y\)

Use the Desmos graph below with the settings \(f(x)=x^2,\,h=3,\,k=0,\,p=-4\)

\((x - 3)^{2} = -16y\)
Vertex \((3, 0)\)
Focus \((3, -4)\)
Directrix \(y = 4\)
Endpoints of latus rectum \((-5, -4)\), \((11, -4)\)

\(\left (x + \frac {7}{3}\right )^{2} = 2\left (y + \frac {5}{2}\right )\)

Use the Desmos graph below with the settings \(f(x)=x^2,\,h=-\frac {7}{3},\,k=-\frac {5}{2},\,p=\frac {1}{2}\)

\(\left (x + \frac {7}{3}\right )^{2} = 2\left (y + \frac {5}{2}\right )\)
Vertex \(\left (-\frac {7}{3}, -\frac {5}{2} \right )\)
Focus \(\left (-\frac {7}{3}, -2 \right )\)
Directrix \(y = -3\)
Endpoints of latus rectum \(\left (-\frac {10}{3}, -2 \right )\), \(\left (-\frac {4}{3}, -2 \right )\)

\((y - 2)^{2} = -12(x + 3)\)

Use the Desmos graph below with the settings \(f(y)=y^2,\,h=-3,\,k=2,\,p=-3\)

\((y - 2)^{2} = -12(x + 3)\)
Vertex \((-3, 2)\)
Focus \((-6, 2)\)
Directrix \(x = 0\)
Endpoints of latus rectum \((-6, 8)\), \((-6, -4)\)

\((y + 4)^{2} = 4x\)

Use the Desmos graph below with the settings \(f(y)=y^2,\,h=0,\,k=-4,\,p=1\)

\((y + 4)^{2} = 4x\)
Vertex \((0,-4)\)
Focus \((1,-4)\)
Directrix \(x = -1\)
Endpoints of latus rectum \((1, -2)\), \((1, -6)\)

\((x-1)^2 = 4(y+3)\)

Use the Desmos graph below with the settings \(f(x)=x^2,\,h=1,\,k=-3,\,p=1\)

\((x-1)^2 = 4(y+3)\)
Vertex \(\left (1, -3\right )\)
Focus \(\left (1, -2 \right )\)
Directrix \(y = -4\)
Endpoints of latus rectum \(\left (3, -2 \right )\), \(\left (-1, -2 \right )\)

\((x+2)^2 = -20(y-5)\)

Use the Desmos graph below with the settings \(f(x)=x^2,\,h=-2,\,k=5,\,p=-5\)

\((x+2)^2 = -20(y-5)\)
Vertex \(\left (-2, 5\right )\)
Focus \(\left (-2, 0 \right )\)
Directrix \(y = 10\)
Endpoints of latus rectum \(\left (-12, 0 \right )\), \(\left (8, 0 \right )\)

\((y-4)^2 = 18(x-2)\)

Use the Desmos graph below with the settings \(f(y)=y^2,\,h=2,\,k=4,\,p=\frac {9}{2}\)

\((y-4)^2 = 18(x-2)\)
Vertex \(\left (2, 4\right )\)
Focus \(\left ( \frac {13}{2}, 4 \right )\)
Directrix \(x = -\frac {5}{2}\)
Endpoints of latus rectum \(\left (\frac {13}{2}, -5 \right )\), \(\left (\frac {13}{2}, 13 \right )\)

\(\left (y+ \frac {3}{2}\right )^2 = -7 \left (x+ \frac {9}{2}\right )\)

Use the Desmos graph below with the settings \(f(y)=y^2,\,h=-\frac {9}{2},\,k=-\frac {3}{2},\,p=-\frac {7}{4}\)

\(\left (y+ \frac {3}{2}\right )^2 = -7 \left (x+ \frac {9}{2}\right )\)
Vertex \(\left (-\frac {9}{2}, -\frac {3}{2}\right )\)
Focus \(\left ( -\frac {25}{4}, -\frac {3}{2} \right )\)
Directrix \(x = -\frac {11}{4}\)
Endpoints of latus rectum \(\left (-\frac {25}{4}, 2 \right )\), \(\left (-\frac {25}{4}, -5 \right )\)

In Exercises stdfrmparabolafirst - stdfrmparabolalast, put the equation into standard form. Find the vertex, focus and directrix.

\(y^{2} - 10y - 27x + 133 = 0\)

\((y - 5)^{2} = 27(x - 4)\)
Vertex \((4, 5)\)
Focus \(\left ( \frac {43}{4}, 5 \right )\)
Directrix \(x = -\frac {11}{4}\)
\(25x^{2} + 20x + 5y - 1 = 0\)

\(\left (x + \frac {2}{5} \right )^{2} = -\frac {1}{5}(y - 1)\)
Vertex \(\left ( -\frac {2}{5}, 1 \right )\)
Focus \(\left ( -\frac {2}{5}, \frac {19}{20} \right )\)
Directrix \(y = \frac {21}{20}\)
\(x^2 + 2x - 8y + 49 = 0\)

\((x+1)^2=8(y-6)\)
Vertex \(\answer {(-1,6)}\)
Focus \(\answer {(-1,8)}\)
Directrix \(\answer {y=4}\)
\(2y^2 + 4y +x - 8 = 0\)

\((y+1)^2=-\frac {1}{2}(x-10)\)
Vertex \((10,-1)\)
Focus \(\left (\frac {79}{8}, -1 \right )\)
Directrix \(x = \frac {81}{8}\)
\(x^2-10x+12y+1=0\)

\((x-5)^2 = -12(y-2)\)
Vertex \(\answer {(5,2)}\)
Focus \(\answer {(5,-1)}\)
Directrix \(\answer {y=5}\)
\(3y^2-27y+4x+\frac {211}{4} = 0\)

\(\left (y-\frac {9}{2}\right )^2 = -\frac {4}{3} (x-2)\)
Vertex \(\left (2, \frac {9}{2}\right )\)
Focus \(\left (\frac {5}{3}, \frac {9}{2}\right )\)
Directrix \(x = \frac {7}{3}\)
For each of the equations given in Exercises parabolasketchfirst - stdfrmparabolalast that do not describe \(y\) as a function of \(x\), find two or more explicit functions of \(x\) represented by each of the equations. (See Example horizontalparabolaex.)

The equations which do not represent \(y\) as a function of \(x\) are: paranotfcnone, paranotfcntwo, paranotfcnthree, paranotfcnfour, paranotfcnfive, paranotfcnsix, paranotfcnseven.

For number paranotfcnone:

  • \(f(x) = 2+2 \sqrt {-3x-9}\) represents the upper half of the parabola.
  • \(g(x) = 2 - 2\sqrt {-3x-9}\) represents the lower half of the parabola.

For number paranotfcntwo:

  • \(f(x) = -4+2\sqrt {x}\) represents the upper half of the parabola.
  • \(g(x) = -4 - 2\sqrt {x}\) represents the lower half of the parabola.

For number paranotfcnthree:

  • \(f(x) =4+3 \sqrt {2x-4}\) represents the upper half of the parabola.
  • \(g(x) = 4-3 \sqrt {2x-4}\) represents the lower half of the parabola.

For number paranotfcnfour:

  • \(f(x) =-\frac {3}{2} + \frac {1}{2} \sqrt {-28x - 126}\) represents the upper half of the parabola.
  • \(g(x) =-\frac {3}{2} - \frac {1}{2} \sqrt {-28x - 126}\) represents the lower half of the parabola.

For number paranotfcnfive:

  • \(f(x) =5+3 \sqrt {3x-12}\) represents the upper half of the parabola.
  • \(g(x) =5- 3 \sqrt {3x-12}\) represents the lower half of the parabola.

For number paranotfcnsix:

  • \(f(x) =-1 + \frac {1}{2} \sqrt {-2x+20}\) represents the upper half of the parabola.
  • \(g(x) =-1 - \frac {1}{2} \sqrt {-2x+20}\) represents the lower half of the parabola.

For number paranotfcnseven:

  • \(f(x) = \frac {9}{2} + \frac {2}{3} \sqrt {-3x+6}\) represents the upper half of the parabola.
  • \(f(x) = \frac {9}{2} - \frac {2}{3} \sqrt {-3x+6}\) represents the lower half of the parabola.

In Exercises buildparafromgraphfirst - buildparafromgraphlast, find an equation for the parabola whose graph is given.

[Picture]

\((x+2)^2 = \frac {1}{2} (y+6)\)
[Picture]

\(\answer {x^2 = -\frac {1}{2}(y-4)}\)

[Picture]

\((y-2)^2=x+4\)

[Picture]

\(\answer {y^2 = -16(x-1)}\)

In Exercises buildparafirst - buildparalast, find an equation for the parabola which fits the given criteria.

Vertex \((7, 0)\), focus \((0, 0)\).

\(y^{2} = -28(x - 7)\)
Focus \((10, 1)\), directrix \(x = 5\).

\((y - 1)^{2} = 10\left (x - \frac {15}{2} \right )\)
Vertex \((-8, -9)\); \((0, 0)\) and \((-16, 0)\) are points on the curve.

\((x + 8)^{2} = \frac {64}{9}(y + 9)\)
The endpoints of latus rectum are \((-2, -7)\) and \((4, -7)\).

\((x - 1)^{2} = 6\left (y + \frac {17}{2}\right )\) or \((x - 1)^{2} = -6\left (y + \frac {11}{2}\right )\)
The mirror in Carl’s flashlight is a paraboloid of revolution. If the mirror is 5 centimeters in diameter and 2.5 centimeters deep, where should the light bulb be placed so it is at the focus of the mirror?

The bulb should be placed \(0.625\) centimeters above the vertex of the mirror.

A parabolic Wi-Fi antenna is constructed by taking a flat sheet of metal and bending it into a parabolic shape. If the cross section of the antenna is a parabola which is 45 centimeters wide and 25 centimeters deep, where should the receiver be placed to maximize reception?

The receiver be placed \(5.0625\) centimeters from the vertex of the cross section of the antenna.
A parabolic arch is constructed which is 6 feet wide at the base and 9 feet tall in the middle. Find the height of the arch exactly 1 foot in from the base of the arch.

The arch can be modeled by \(x^2=-(y-9)\) or \(y=9-x^2\). One foot in from the base of the arch corresponds to either \(x = \pm 2\), so the height is \(y=9-(\pm 2)^2=5\) feet.
A popular novelty item is the ‘mirage bowl.’ Follow this link to see another startling application of the reflective property of the parabola.
With the help of your classmates, research spinning liquid mirrors. To get you started, here.