how to find equation of parabola with focus and directrix

Parabolas
(This section created by Jack Sarfaty)

Objectives:

Lesson 1: Notice the standard form of a quadratic function, and then find the vertex, line of symmetry, and maximum or minimum value for the defined quadratic role.
Lesson 2: Find the vertex, focus, and directrix, and draw a graph of a parabola, given its equation.
Lesson 3: Find the equation of our parabola when we are given the coordinates of its focus and vertex.
Lesson 4: Find the vertex, focus, and directrix, and graph a parabola by get-go completing the foursquare.

Lesson one

The Parabola is defined equally "the set of all points P in a plane equidistant from a fixed line and a stock-still point in the plane." The stock-still line is called the directrix, and the fixed point is called the focus.

A parabola, as shown on the cables of the Gold Gate Span (beneath), tin can be seen in many different forms. The path that a thrown ball takes or the flow of water from a hose each illustrate the shape of the parabola.

Each parabola is, in some form, a graph of a second-degree function and has many properties that are worthy of examination. Allow's begin by looking at the standard form for the equation of a parabola.

The standard form is (10 - h)^two = 4p (y - k), where the focus is (h, k + p) and the directrix is y = yard - p. If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis, it has an equation of (y - k)² = 4p (x - h), where the focus is (h + p, k) and the directrix is x = h - p.

It would also be in our best involvement to encompass some other class that the equation of a parabola may appear as
y = (ten - h)ⁱⁱ + k, where h represents the distance that the parabola has been translated forth the ten axis, and k represents the distance the parabola has been shifted up and downwardly the y-axis.

Completing the square to get the standard form of a parabola.

We should now determine how nosotros will arrive at an equation in the form y = (x - h)² + yard;

Example 1

Suppose we are given an equation like
y = 3x^two + 12x + 1.

Nosotros now need to complete the square for this equation. I will assume that yous have had some pedagogy on completing the square; but in case y'all haven't, I will go through one instance and get out the rest to the reader.

When completing the square, we first accept to isolate the Ax² term and the By term from the C term. And then the kickoff couple of steps will only deal with the get-go ii parts of the trinomial.

In gild to complete the square, the quadratic in the form y = Ax^two + By + C cannot have an A term that is anything other than 1. In our example, A = 3; so we now need to divide the 3 out, simply that is only out of the 3x² + 12x terms.

This simplifies to y = iii(10² + 4x) + 1. From here we need to take 1/two of our B term, so square the production. So in this case, nosotros have 1/2(4) = 2, then two^two is 4. Now, accept that 4 and place information technology inside the parenthetical term.

To update what we have: y = 3(x^two + 4x + 4) + i; but nosotros at present need to keep in listen that we accept added a term to our equation that must be accounted for. By adding four to the within of the parenthesis, nosotros have done more than only add 4 to the equation. We have at present added 4 times the iii that is sitting in front of the parenthetical term. So, really we are adding 12 to the equation, and we must now offset that on the same side of the equation. We will now offset past subtracting 12 from that one we left off to the correct hand side.

To update: y = 3(x² + 4x + 4) + one - 12. We have now successfully completed the square. At present nosotros need to get this into more than friendly terms. The inside of the parenthesis (the completed foursquare) can be simplified to (x + 2)². The terminal version after the fume clears is y = three(x + 2)² - 11. And , oh the wealth of information we can pull from something like this! Nosotros will find the specifics from this type of equation below.

Finding the vertex, line of symmetry, and maximum and minimum value for the defined quadratic function.

Let'south commencement focus on the second form mentioned, y =(x - h)² + k. When we accept an equation in this grade, we tin can safely say that the 'h' represents the same thing that 'h' represented in the showtime standard form that we mentioned, as does the 'k'. When we have an equation like y = (x - 3)ⁱⁱ + iv, nosotros run across that the graph has been shifted 3 units to the right and iv units upward. The picture below shows this parabola in the start quadrant.

Had the within of the parenthesis in the example equation read,"(x+three)" as opposed to "(x-3)," then the graph would take been shifted three units to the left of the origin. The "+4" at the end of the equation tells the graph to shift upwardly 4 units. Too, had the equation read "-iv," then the graph would still be pointed up, but the vertex would have been 4 units below the ten-axis.

A groovy deal can exist determined by an equation in this course.

The Vertex

The most obvious thing that we can tell, without having to wait at the graph, is the origin. The origin tin can be institute by pairing the h value with the yard value, to give the coordinate (h, chiliad). The almost obvious mistake that can arise from this is past taking the wrong sign of the 'h.' In our case equation, y = (x - 3)² + 4, nosotros noticed that the 'h' is 3, but it is often mistaken that the ten-coordinate of our vertex is -3; this is not the case because our standard grade for the equation is y = (x - h)ⁱⁱ + k, implying that the nosotros demand to change the sign of what is inside the parenthesis.

The Line of Symmetry

To find the line of symmetry of a parabola in this form, we demand to remember that nosotros are simply dealing with parabolas that are pointed up or downwardly in nature. With this in mind, the line of symmetry (also known equally the centrality of symmetry) is the line that splits the parabola into two dissever branches that mirror each other. The line of symmetry goes through the vertex, and since we are now only dealing with parabolas that become upwardly and down, the line of symmetry must be a vertical line that will begin with "x = _ ". The number that goes in this blank will be the x-coordinate of the vertex. For example, when we looked at y = (x - three)² + 4, the x-coordinate of the vertex is going be 3; and then the equation for the line of symmetry is x = 3.

In order to visualize the line of symmetry, accept the picture of the parabola above and draw an imaginary vertical line through the vertex. If yous were to take the equation of that vertical line, y'all would detect that the line is going through the x-axis at x = 3. An piece of cake fault that students often brand is that they say that the line of symmetry is y = 3 since the line is vertical. We must keep in listen that the equations for vertical and horizontal lines are the reverse of what you expect them to be. Nosotros always say that vertical ways "up and downward; so the equation of the line (existence parallel to the y-axis) begins with 'y =__'," but we forget that the key is which centrality the line goes through. So since the line goes through the x-centrality, the equation for this vertical line must be x = __.

The Maximum or Minimum

In the line of symmetry word, nosotros dealt with the ten-coordinate of the vertex; and only like clockwork, we demand to now examine the y-coordinate. The y-coordinate of the vertex tells u.s.a. how high or how low the parabola sits.

One time again with our trusty example, y = (x-3)² + 4, we see that the y-coordinate of the vertex (as derived from the number on the far right of the equation) dictates how loftier or low on the coordinate plane that the parabola sits. This parabola is resting on the line y = 4 (see line of symmetry for why the equation is y = __, instead of ten = __ ). Once nosotros have identified what the y-coordinate is, the last question we have is whether this number represents a maximum or minimum. We call this number a maximum if the parabola is facing down (the vertex represents the highest indicate on the parabola), and we can call it a minimum if the parabola is facing upward (the vertex represents the lowest point on the parabola).

How do we tell if the parabola is pointed up or downward by just looking at the equation?

Every bit long as nosotros accept the equation in the form derived from the completing the square footstep, we look and come across if there is a negative sign in front of the parenthetical term. If the equation comes in the class of y = - (x - h)² + k, the negative in front of the parenthesis tells us that the parabola is pointed down (as illustrated in the moving picture below). If there is no negative sign in front, so the parabola faces upward.

Instance 2

Let's now look at an example of another equation of a parabola in standard form. Nosotros will then place its vertex, line of symmetry, and maximum or minimum.

For example, allow's take the equation y = - (x + 4)² - 7. The outset thing we would similar to do is look at the graph of the bend. This should help us make sense of the things we are looking for. The graph is shown below.

As you can run across, this bend falls into the third quadrant and is pointed downward. The vertex appears to accept a negative x-coordinate and a negative y-coordinate. We will look more closely at the equation and take what we accept already learned, we should exist satisfied with our results.

y = - (x + 4)^two - seven gives us a vertex of (-iv, -seven). The x-coordinate is the "reverse" of 4, which is -4, and the y-coordinate is -7 as seen from the number that sits at the end of the equation.

Beginning, the negative sign at the showtime of the equation immediately tips us off that the parabola is facing downwardly.

Next, the ten-coordinate that we found is the primal to finding the line of symmetry. We know that the equation for the line of symmetry volition be "x = __ ," and the number inside the bare is the x-coordinate, -iv.

Lastly, we need to decide whether nosotros have a maximum or minimum. The y-coordinate is going to be a maximum in this example considering the vertex lies on the highest indicate (maximum) of our curve. Then in this example, we have a maximum of y = -7.

Let'due south now await at the same curve above with the vertex, line of symmetry, and maximum visible:

Plainly, the ruby-red bend represents the parabola. The green line represents our line of symmetry (equation x = -4) and the blueish line represents the line that the maximum rests on at y = -7. Hopefully this visual has helped y'all run across all of the specific parts that we take discussed and then far.

For some supplementary exercises over what nosotros have covered so far click here: Practise i

(Back to tiptop)

Lesson ii

Detect the vertex, focus, and directrix, and draw a graph of a parabola, given its equation.

As yous may or may not know, a parabola is the locus of points in a plane equidistant from a fixed line and a fixed signal on the plane. Nosotros know this stock-still line to exist the directrix and the fixed point to be the focus.

To meet an animated picture of the above description, y'all need to have Geometer's SketchPad for either Macintosh or PC loaded on your computer. If you have GSP, click here. To download the script of this flick so y'all can create it yourself, click hither.

Let's now take a look at a parabola that has all of the elements that we will be looking for:

the vertex
the focus
the directrix

The following example is especially meant for those who do not have GSP on your computer. This picture (below) is generated from Algebra Xpresser.

From the above picture, I have labeled 3 items that we need to pay close attending to. The highest point of the parabola is the vertex (and the maximum). The plus sign that is straight under the vertex is the focus. The light-green line that is higher up the parabola (and direct above the vertex) is the directrix. You may be able to see, past eyeballing, that the distance from the focus to the vertex is the aforementioned distance as the vertex to the directrix. We will now go into a chip of particular as to how to derive all of this information from a given equation.

The next example that I will give you will be a dainty, easy equation from which nosotros can easily option the information we need.

Example three

Let's examine the equation, (ten + 2)² = -6(y - 1).

Patently, this equation is different from the "vertex class" we learned in the prior lesson. Even still we can discover all of the information we institute in the first lesson: the vertex, line of symmetry, and the maximum/minimum. We take to use the same line of thought that the vertex is where the x and y terms are. In the same way nosotros observe the x-coordinate is -2, the y-coordinate is 1 {Five: (-2, 1)}. All we need to find the line of symmetry and the maximum/minimum is the vertex; and then permit's follow through: The line of symmetry is x = -2, and the maximum (since we accept a negative sign in front end of 1 of our terms) is at y = 1.

Now for the fun part.

In order to find the focus and directrix of the parabola, we need to have the equations that give an upward or down facing parabola in the form (x - h)² = 4p(y - k) form. In other words, we demand to have the x² term isolated from the rest of the equation. We are used to having ten^two by itself, just if the vertex has been shifted either upwardly or down, we demand to show this in the parenthetical term with the y. The coefficient of the (y - grand) term is the 4p term. We demand to take this number and set up it equal to 4p.

In this example, 4p is equal to the term in forepart of the y term (in parenthesis); then 4p = -6. This ways that p = -3/two. Since this is an downward facing parabola, we need to have the focus inside of the curve, meaning the focus is below the vertex. How far below the vertex? Take the y-coordinate and add together the p term information technology. And so, nosotros now accept the vertex at (-two, 1) and nosotros are, in essence, subtracting -3/2 from one. This will move the focus to the point (-ii, -1/two).

The directrix is equidistant from the vertex that the focus is. And then if the focus is downwards -3/two from the vertex, then the directrix is a line that is upward 3/2 from the vertex. That puts the directrix at y = 5/2.

For an illustration of this problem, delight look at the moving picture below:

The greenish line (three/ii units up from the vertex) is the directrix, and the plus sign 3/2 units downwardly from the vertex is the focus.

This should assist united states of america with the parabolas that open up upwards and downward. Let's now take a look at a parabolat that opens left and right.

Case 4

Let's take a wait at the equation (y + iii)²= 12 (x - 1).

We tin can easily identify that the parabola is opening left or right. Since the coefficient in front end of the ten term is positive, we can say that the parabola volition open to the correct. The focus will be to the right of the vertex, and the directrix will exist a vertical line that is the aforementioned distance to the left of the vertex that the focus is to the correct.

The vertex is (1, -three), the axis of symmetry (now horizontal) is y = -3, and nosotros don't recognize "max'southward and min's" for parabolas that open left or right.

The term in front of the x term is a 12. This is what our 4p term is equal to. And so 4p = 12, making p = 3. So we at present need to move the focus 3 units right from the the origin. This ways that the coordinate for the focus is (four, -3), and the directrix will exist a vertical line going through the point (-ii, -three).

This problem is illustrated in the motion picture below.

Once over again, our green line represents the directrix and the plus sign represents the aforementioned focus.

For some supplementary exercises over what we have covered in lesson ii, click hither: Exercise two

(Back to top)

Lesson three

Discover the equation of a parabola when we are given the coordinates of its focus and vertex.

Now, we are going to begin taking what we have learned and first piecing it together. If nosotros are given a focus and a vertex, we have plenty to exist able to generate a quadratic equation of a parabola. If we think well-nigh it for a second, nosotros volition be able to find the altitude from the vertex to the focus based on this given data. We will then be able to calculate our p term (the term from the previous lesson that is in front end of our non-squared variable). Placing the coordinates of the vertex into the equation is very elementary, relative to what we take learned so far.

Example 5

Let us suppose that we are given a focus of (-vi, 0) and the vertex is at the origin.

Based on what nosotros know without plugging anything in, nosotros can say that the parabola will exist opening upwards to the left because its focus is to the left of the origin. Now in beginning to slice things together, nosotros tin can say that the equation volition be something like yⁱⁱ is equal to some 10 term.

Since the origin is the vertex, nosotros can say that this will exist (y - 0)² = 4p(x - 0), which simplifies to y² = 4px.

We know that p = -half dozen, and we know that 4p = -24. We should at present be able to tell that the equation is y² = -24x.

Instance 6

We will at present try a problem that has the parabola opening upwardly or downward. Nosotros will make the focus (2, 3) and the vertex (2, 6).

The focus is straight below the vertex by iii units; so p = -3; then 4p = -12; but not so fast! Nosotros aren't quite home gratuitous yet. The vertex is shifted off of the origin, and nosotros need to consider the h and k terms.

The equation with a parabola facing downwards will exist (x - h)² = 4p(y - thousand), where 4p is negative. To again piece things together:
(10 - ii)² = -12 (y - ii).

For some supplementary exercises over what we take covered in lesson 3, click hither: Exercise 3

(Dorsum to meridian)

Lesson 4

Find the vertex, focus, and directrix, and graph a parabola by first completing the foursquare.

Not e'er exercise nosotros come up up on equations that are there just waiting for the states to solve them. Sometimes we've got to work a bit to discover their primal points. Hopefully this example will lead the states toward such a trouble.

Example vii

The last pair of examples that we volition examine will be one where we are given a quadratic equation that is not already in any item standard grade.
We will now be forced to consummate the square to arrive at the form we need to find the newest parts of the parabola that nosotros have explored.

Suppose that we have x² + 6x +4y + 5 = 0. Since the x-term is the squared term, nosotros will choose to isolate all of the terms that have ten in them. We volition need to identify the x terms on one side of the equation, while the rest of the terms are on the opposite side.

This footstep volition exit us with x² + 6x = -4y - 5. When nosotros complete the square on the left-mitt side of the equation, we will have ten² + 6x + 9; sowe will need to add ix to the right-hand side, as well.

This will bring us to (x + 3)² = -4y + 4. Remembering that any coefficients of the x or y terms demand to get in front of the non-squared variable, we volition factor the -4 from the y-term. This volition leave us with (x + iii)^two = -4 (y - 1).

From here, the trouble resembles both of the others.

Our vertex is (-3, 1), our line of symmetry is x = -3; and we practice have a maximum at y = ane;

The focus can now be found by taking the number in front end of the non-squared variable -4 and setting it equal to 4p. 4p = -four; so p = -1.

Since the parabola is facing downward, the focus is beneath the vertex, and the directrix is above. We will accept our vertex and add (-1) to the y-coordinate. This will accept united states to the point (-iii, 0) that is our focus. The directrix (on the opposite side of the vertex) is at the horizontal line y = 2. Once again, we volition look at an analogy below. The greenish line is the directrix, and blueish dot is the focus.

(Corrections by J. Wilson, 28 February 2012)

Example 8

Suppose that we accept an equation y² + 2y + 4x -8 = 0. Before nosotros doing whatever steps, such as completing the square, we tin can see that the squared term is the y term. This will tell u.s. that the parabola with either open to the left or to the right. Since this is the case where the y term is squared, nosotros need to isolate the y terms to one side of the equation and put the x terms and the constants on the other. We will stick to what we've been doing all along and put the isolated terms on the left.

Subsequently the isolation stride, we encounter that nosotros take y^two + 2y = -4x + 8.

In completing the square, we will accept y + 2y + i = -4x + eight + 1; this simplifies to (y + 1)² = -4x + 9.

No matter how ugly the right-paw side of the equation may become, nosotros need to divide the right mitt side by the coefficient of the 10 term (in this instance, -iv). This will leave us with (y + 1)² = -4(x - 9/4). From hither nosotros tin say that the parabola volition open up to the left.

We can at present run across that the vertex will be at (9/4, -1).

The term in front of the x is a -iv. This is our 4p value. So we at present tin say that 4p = -iv. In plough, our p = -ane.

Now will we determine that the focus is one unit left of the vertex; then the focus (after some work with fractions) is (v/4, -1).

The directrix is going to be a vertical line that is one unit to the correct of the vertex. So the directrix will be a line, x = 13/4.

Beneath, we will see the sketch of this equation.

Equally has been the case so far, the plus sign represents the focus, which resides at the point (v/4, -1); the directrix is represented by the green line, which is on the equation 10 = 13/4.

For some supplementary exercises over what nosotros have covered in lesson four, click hither: Exercise iv

Source: http://jwilson.coe.uga.edu/EMT725/Class/Sarfaty/EMT669/InstructionalUnit/Parabolas/parabolas.html

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