# 10.3 The parabola  (Page 7/11)

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${\left(y+4\right)}^{2}=16\left(x+4\right)$

${y}^{2}+12x-6y+21=0$

${\left(y-3\right)}^{2}=-12\left(x+1\right),V:\left(-1,3\right);F:\left(-4,3\right);d:x=2$

${x}^{2}-4x-24y+28=0$

$5{x}^{2}-50x-4y+113=0$

${\left(x-5\right)}^{2}=\frac{4}{5}\left(y+3\right),V:\left(5,-3\right);F:\left(5,-\frac{14}{5}\right);d:y=-\frac{16}{5}$

${y}^{2}-24x+4y-68=0$

${x}^{2}-4x+2y-6=0$

${\left(x-2\right)}^{2}=-2\left(y-5\right),V:\left(2,5\right);F:\left(2,\frac{9}{2}\right);d:y=\frac{11}{2}$

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

$3{y}^{2}-4x-6y+23=0$

${\left(y-1\right)}^{2}=\frac{4}{3}\left(x-5\right),V:\left(5,1\right);F:\left(\frac{16}{3},1\right);d:x=\frac{14}{3}$

${x}^{2}+4x+8y-4=0$

## Graphical

For the following exercises, graph the parabola, labeling the focus and the directrix.

$x=\frac{1}{8}{y}^{2}$

$y=36{x}^{2}$

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

$y=-9{x}^{2}$

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

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

$-6{\left(y+5\right)}^{2}=4\left(x-4\right)$

${y}^{2}-6y-8x+1=0$

${x}^{2}+8x+4y+20=0$

$3{x}^{2}+30x-4y+95=0$

${y}^{2}-8x+10y+9=0$

${x}^{2}+4x+2y+2=0$

${y}^{2}+2y-12x+61=0$

$-2{x}^{2}+8x-4y-24=0$

For the following exercises, find the equation of the parabola given information about its graph.

Vertex is $\text{\hspace{0.17em}}\left(0,0\right);$ directrix is $\text{\hspace{0.17em}}y=4,$ focus is $\text{\hspace{0.17em}}\left(0,-4\right).$

${x}^{2}=-16y$

Vertex is $\text{\hspace{0.17em}}\left(0,0\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}x=4,$ focus is $\text{\hspace{0.17em}}\left(-4,0\right).$

Vertex is $\text{\hspace{0.17em}}\left(2,2\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}x=2-\sqrt{2},$ focus is $\text{\hspace{0.17em}}\left(2+\sqrt{2},2\right).$

${\left(y-2\right)}^{2}=4\sqrt{2}\left(x-2\right)$

Vertex is $\text{\hspace{0.17em}}\left(-2,3\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}x=-\frac{7}{2},$ focus is $\text{\hspace{0.17em}}\left(-\frac{1}{2},3\right).$

Vertex is $\text{\hspace{0.17em}}\left(\sqrt{2},-\sqrt{3}\right);$ directrix is $\text{\hspace{0.17em}}x=2\sqrt{2},$ focus is $\text{\hspace{0.17em}}\left(0,-\sqrt{3}\right).$

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

Vertex is $\text{\hspace{0.17em}}\left(1,2\right);\text{\hspace{0.17em}}$ directrix is $\text{\hspace{0.17em}}y=\frac{11}{3},$ focus is $\text{\hspace{0.17em}}\left(1,\frac{1}{3}\right).$

For the following exercises, determine the equation for the parabola from its graph.

${x}^{2}=y$

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

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

## Extensions

For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

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

${\left(y+1\right)}^{2}=12\left(x+3\right)$

## Real-world applications

The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as $\text{\hspace{0.17em}}{x}^{2}=4y.\text{\hspace{0.17em}}$ At what coordinates should you place the light bulb?

$\left(0,1\right)$

If we want to construct the mirror from the previous exercise such that the focus is located at $\text{\hspace{0.17em}}\left(0,0.25\right),$ what should the equation of the parabola be?

A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

At the point 2.25 feet above the vertex.

Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

0.5625 feet

If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth.

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

${x}^{2}=-125\left(y-20\right),$ height is 7.2 feet

If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet.

An object is projected so as to follow a parabolic path given by $\text{\hspace{0.17em}}y=-{x}^{2}+96x,$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the horizontal distance traveled in feet and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the height. Determine the maximum height the object reaches.

2304 feet

For the object from the previous exercise, assume the path followed is given by $\text{\hspace{0.17em}}y=-0.5{x}^{2}+80x.\text{\hspace{0.17em}}$ Determine how far along the horizontal the object traveled to reach maximum height.

#### Questions & Answers

A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim
Is there any rule we can use to get the nth term ?