How To Find The Directrix Of A Parabola

Parabolas
(This section created by Jack Sarfaty)

Objectives:

Lesson 1: Detect the standard form of a quadratic part, and and then discover the vertex, line of symmetry, and maximum or minimum value for the defined quadratic function.
Lesson 2: Find the vertex, focus, and directrix, and describe 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 first completing the square.

Lesson i

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

A parabola, equally shown on the cables of the Golden Gate Span (below), can exist seen in many different forms. The path that a thrown ball takes or the menses of h2o from a hose each illustrate the shape of the parabola.

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

The standard form is (x - h)² = 4p (y - one thousand), where the focus is (h, grand + p) and the directrix is y = m - 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 - chiliad)² = 4p (10 - h), where the focus is (h + p, k) and the directrix is x = h - p.

It would also be in our best interest to embrace another form that the equation of a parabola may appear as
y = (x - h)² + k, where h represents the altitude that the parabola has been translated along the x centrality, and 1000 represents the distance the parabola has been shifted up and downward the y-axis.

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

We should now determine how we volition get in at an equation in the course y = (x - h)² + 1000;

Case ane

Suppose we are given an equation like
y = 3x² + 12x + i.

Nosotros now need to complete the square for this equation. I volition presume that you have had some instruction on completing the foursquare; but in instance you haven't, I will become through one example and go out the rest to the reader.

When completing the square, we first have to isolate the Ax² term and the By term from the C term. So the first couple of steps will only deal with the first two parts of the trinomial.

In order to complete the foursquare, the quadratic in the course y = Ax^two + By + C cannot have an A term that is anything other than ane. In our example, A = 3; then we at present need to separate the 3 out, but that is merely out of the 3x² + 12x terms.

This simplifies to y = 3(10² + 4x) + i. From here we demand to accept one/2 of our B term, then square the product. So in this case, we accept 1/2(4) = 2, then ii^two is 4. Now, take that 4 and place information technology inside the parenthetical term.

To update what nosotros accept: y = three(x^two + 4x + 4) + 1; just we now need to keep in mind that we have added a term to our equation that must be accounted for. By adding 4 to the inside of the parenthesis, we have washed more than just add 4 to the equation. We accept now added iv times the 3 that is sitting in front end of the parenthetical term. So, really we are adding 12 to the equation, and we must now get-go that on the same side of the equation. We volition now kickoff past subtracting 12 from that ane we left off to the right hand side.

To update: y = 3(xⁱⁱ + 4x + 4) + 1 - 12. Nosotros have at present successfully completed the foursquare. Now we need to get this into more friendly terms. The inside of the parenthesis (the completed foursquare) can exist simplified to (ten + 2)ⁱⁱ. The last version afterwards the smoke clears is y = iii(x + 2)² - xi. And , oh the wealth of data we tin pull from something like this! We will notice the specifics from this type of equation below.

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

Let's first focus on the 2d form mentioned, y =(ten - h)² + g. When we have an equation in this class, we tin can safely say that the 'h' represents the aforementioned thing that 'h' represented in the get-go standard course that nosotros mentioned, as does the 'k'. When we take an equation like y = (x - 3)^two + 4, nosotros run into that the graph has been shifted 3 units to the right and iv units upwardly. The picture below shows this parabola in the first quadrant.

Had the inside of the parenthesis in the example equation read,"(x+three)" as opposed to "(x-3)," and so the graph would have been shifted three units to the left of the origin. The "+4" at the end of the equation tells the graph to shift up 4 units. Likewise, had the equation read "-iv," and so the graph would still be pointed upwards, but the vertex would take been 4 units beneath the x-centrality.

A corking deal can exist determined by an equation in this class.

The Vertex

The well-nigh obvious matter that nosotros tin tell, without having to look at the graph, is the origin. The origin tin be plant by pairing the h value with the chiliad value, to give the coordinate (h, k). The nearly obvious fault that can arise from this is by taking the wrong sign of the 'h.' In our example equation, y = (x - 3)^two + iv, we noticed that the 'h' is 3, but it is often mistaken that the ten-coordinate of our vertex is -iii; this is not the case because our standard course for the equation is y = (ten - h)² + k, implying that the we need to alter the sign of what is within the parenthesis.

The Line of Symmetry

To find the line of symmetry of a parabola in this form, nosotros demand to remember that nosotros are simply dealing with parabolas that are pointed up or down in nature. With this in mind, the line of symmetry (too known equally the axis of symmetry) is the line that splits the parabola into two separate branches that mirror each other. The line of symmetry goes through the vertex, and since nosotros are at present simply dealing with parabolas that get up and down, the line of symmetry must be a vertical line that will brainstorm with "x = _ ". The number that goes in this bare will be the 10-coordinate of the vertex. For case, when nosotros looked at y = (x - three)^two + 4, the 10-coordinate of the vertex is going be 3; and so the equation for the line of symmetry is ten = three.

In guild to visualize the line of symmetry, take the picture of the parabola above and describe an imaginary vertical line through the vertex. If you were to take the equation of that vertical line, you would discover that the line is going through the ten-axis at 10 = iii. An easy mistake that students often make is that they say that the line of symmetry is y = 3 since the line is vertical. We must keep in mind that the equations for vertical and horizontal lines are the reverse of what you expect them to be. Nosotros always say that vertical means "up and down; so the equation of the line (being parallel to the y-axis) begins with 'y =__'," but we forget that the key is which axis the line goes through. So since the line goes through the 10-axis, the equation for this vertical line must exist 10 = __.

The Maximum or Minimum

In the line of symmetry discussion, we dealt with the x-coordinate of the vertex; and just like clockwork, nosotros need to at present examine the y-coordinate. The y-coordinate of the vertex tells usa how loftier or how depression the parabola sits.

Over again with our trusty instance, y = (x-three)ⁱⁱ + iv, we run into that the y-coordinate of the vertex (equally derived from the number on the far right of the equation) dictates how high or low on the coordinate plane that the parabola sits. This parabola is resting on the line y = 4 (come across line of symmetry for why the equation is y = __, instead of x = __ ). Once we 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 downward (the vertex represents the highest point on the parabola), and we can phone call it a minimum if the parabola is facing upwards (the vertex represents the lowest signal on the parabola).

How practise we tell if the parabola is pointed upward or downwards by just looking at the equation?

As long as we take the equation in the form derived from the completing the foursquare stride, we look and see if there is a negative sign in forepart of the parenthetical term. If the equation comes in the form of y = - (x - h)² + k, the negative in front of the parenthesis tells us that the parabola is pointed downward (every bit illustrated in the picture show below). If at that place is no negative sign in front end, then the parabola faces upwardly.

Instance 2

Let's at present await at an example of another equation of a parabola in standard form. We volition then identify its vertex, line of symmetry, and maximum or minimum.

For example, let'due south take the equation y = - (x + 4)² - 7. The first thing we would like to do is look at the graph of the curve. This should help us brand sense of the things nosotros are looking for. The graph is shown below.

As you tin can see, this curve falls into the third quadrant and is pointed downwardly. The vertex appears to have a negative 10-coordinate and a negative y-coordinate. We will look more than closely at the equation and take what nosotros take already learned, we should be satisfied with our results.

y = - (x + 4)ⁱⁱ - vii gives united states of america a vertex of (-4, -seven). The x-coordinate is the "contrary" of four, which is -four, 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 outset of the equation immediately tips us off that the parabola is facing downward.

Next, the ten-coordinate that nosotros found is the key to finding the line of symmetry. We know that the equation for the line of symmetry will be "ten = __ ," and the number inside the blank is the x-coordinate, -4.

Lastly, we need to decide whether nosotros have a maximum or minimum. The y-coordinate is going to exist a maximum in this case because the vertex lies on the highest point (maximum) of our curve. So in this case, we accept a maximum of y = -vii.

Let'southward now await at the aforementioned bend above with the vertex, line of symmetry, and maximum visible:

Evidently, the red bend represents the parabola. The green line represents our line of symmetry (equation x = -iv) and the blueish line represents the line that the maximum rests on at y = -7. Hopefully this visual has helped you see all of the specific parts that we accept discussed and then far.

For some supplementary exercises over what we have covered and then far click here: Exercise i

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Lesson 2

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

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

To see an animated picture of the above description, you lot need to accept Geometer's SketchPad for either Macintosh or PC loaded on your reckoner. If you lot accept GSP, click hither. To download the script of this picture and then you tin can create it yourself, click here.

Let'due south now have a look at a parabola that has all of the elements that nosotros will be looking for:

the vertex
the focus
the directrix

The post-obit example is especially meant for those who exercise non have GSP on your reckoner. This picture (below) is generated from Algebra Xpresser.

From the above movie, I have labeled 3 items that nosotros need to pay close attention 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 green line that is above the parabola (and direct above the vertex) is the directrix. You may be able to see, by 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 bit of detail equally to how to derive all of this information from a given equation.

The next instance that I will give you volition be a nice, easy equation from which we can easily pick the data we need.

Example 3

Let's examine the equation, (x + two)² = -six(y - 1).

Plain, this equation is unlike from the "vertex form" we learned in the prior lesson. Even still we tin can find all of the information we found in the first lesson: the vertex, line of symmetry, and the maximum/minimum. Nosotros take to utilise the same line of thought that the vertex is where the x and y terms are. In the same manner we find the x-coordinate is -2, the y-coordinate is ane {V: (-2, 1)}. All we need to discover the line of symmetry and the maximum/minimum is the vertex; then let'due south follow through: The line of symmetry is x = -2, and the maximum (since nosotros accept a negative sign in front end of one of our terms) is at y = i.

Now for the fun part.

In order to find the focus and directrix of the parabola, we demand to have the equations that give an upwardly or downwardly facing parabola in the form (ten - h)² = 4p(y - grand) form. In other words, we need to have the 10^two term isolated from the residual of the equation. We are used to having x² by itself, only if the vertex has been shifted either up or down, we need to bear witness this in the parenthetical term with the y. The coefficient of the (y - k) term is the 4p term. We need to take this number and set it equal to 4p.

In this case, 4p is equal to the term in front of the y term (in parenthesis); so 4p = -6. This means that p = -3/2. Since this is an downwards facing parabola, we need to take the focus inside of the curve, meaning the focus is beneath the vertex. How far below the vertex? Take the y-coordinate and add the p term it. So, we now take the vertex at (-ii, 1) and we are, in essence, subtracting -iii/2 from ane. This will move the focus to the point (-2, -1/2).

The directrix is equidistant from the vertex that the focus is. Then if the focus is downward -three/ii from the vertex, and then the directrix is a line that is up three/2 from the vertex. That puts the directrix at y = 5/2.

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

The green line (three/ii units up from the vertex) is the directrix, and the plus sign 3/two units down from the vertex is the focus.

This should aid us with the parabolas that open upward and downwardly. Let'southward at present take a wait at a parabolat that opens left and right.

Example 4

Let's take a look at the equation (y + three)²= 12 (10 - 1).

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

The vertex is (one, -3), the centrality of symmetry (at present horizontal) is y = -3, and we don't recognize "max'due south and min's" for parabolas that open left or right.

The term in forepart of the ten term is a 12. This is what our 4p term is equal to. So 4p = 12, making p = three. So we now need to move the focus three units correct from the the origin. This means that the coordinate for the focus is (4, -3), and the directrix volition be a vertical line going through the point (-2, -3).

This trouble is illustrated in the picture below.

One time once again, our green line represents the directrix and the plus sign represents the aforementioned focus.

For some supplementary exercises over what we accept covered in lesson 2, click here: Exercise 2

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Lesson iii

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

Now, we are going to brainstorm taking what we accept learned and start piecing it together. If we are given a focus and a vertex, we have enough to be able to generate a quadratic equation of a parabola. If nosotros retrieve about it for a 2nd, nosotros volition exist able to find the distance from the vertex to the focus based on this given information. Nosotros will then be able to calculate our p term (the term from the previous lesson that is in front of our not-squared variable). Placing the coordinates of the vertex into the equation is very unproblematic, relative to what we have learned so far.

Instance five

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

Based on what we know without plugging anything in, we can say that the parabola will be opening up to the left because its focus is to the left of the origin. Now in first to piece things together, nosotros can say that the equation will be something like y² is equal to some x term.

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

We know that p = -6, and we know that 4p = -24. We should now be able to tell that the equation is y² = -24x.

Example 6

Nosotros will now try a problem that has the parabola opening up or downwardly. We will make the focus (two, 3) and the vertex (2, 6).

The focus is directly beneath the vertex by 3 units; so p = -3; so 4p = -12; just not so fast! We aren't quite home free yet. The vertex is shifted off of the origin, and we need to consider the h and k terms.

The equation with a parabola facing downwardly will exist (x - h)^two = 4p(y - k), where 4p is negative. To over again piece things together:
(x - 2)² = -12 (y - 2).

For some supplementary exercises over what we accept covered in lesson three, click here: Do three

(Back to tiptop)

Lesson 4

Find the vertex, focus, and directrix, and graph a parabola past kickoff completing the square.

Not always do we come up on equations that are in that location but waiting for us to solve them. Sometimes we've got to work a bit to discover their central points. Hopefully this instance will lead us toward such a trouble.

Example vii

The last pair of examples that we will examine volition exist one where we are given a quadratic equation that is non already in whatsoever particular standard form.
Nosotros will at present be forced to consummate the square to make it at the form nosotros need to find the newest parts of the parabola that we take explored.

Suppose that nosotros have ten² + 6x +4y + 5 = 0. Since the x-term is the squared term, we will cull to isolate all of the terms that accept x in them. Nosotros will need to identify the ten terms on one side of the equation, while the rest of the terms are on the opposite side.

This step volition leave us with xⁱⁱ + 6x = -4y - 5. When we consummate the square on the left-paw side of the equation, we volition have 10² + 6x + ix; sowe will need to add 9 to the right-manus side, besides.

This volition bring us to (10 + three)² = -4y + 4. Remembering that any coefficients of the ten or y terms need to go in front of the non-squared variable, we will cistron the -iv from the y-term. This will get out us with (x + 3)² = -4 (y - 1).

From here, the problem resembles both of the others.

Our vertex is (-3, 1), our line of symmetry is x = -three; and nosotros do take a maximum at y = 1;

The focus can now be found by taking the number in forepart of the not-squared variable -4 and setting it equal to 4p. 4p = -4; so p = -ane.

Since the parabola is facing downwards, the focus is below the vertex, and the directrix is higher up. We will accept our vertex and add (-one) to the y-coordinate. This volition have u.s.a. 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. One time again, we volition look at an illustration below. The green line is the directrix, and blueish dot is the focus.

(Corrections past J. Wilson, 28 Feb 2012)

Example eight

Suppose that we have an equation y² + 2y + 4x -8 = 0. Earlier we doing any steps, such as completing the square, we can see that the squared term is the y term. This will tell us that the parabola with either open to the left or to the right. Since this is the case where the y term is squared, we 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 nosotros've been doing all forth and put the isolated terms on the left.

After the isolation step, we encounter that nosotros take yⁱⁱ + 2y = -4x + 8.

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

No matter how ugly the right-hand side of the equation may get, nosotros need to split the right hand side by the coefficient of the x term (in this example, -4). This will leave the states with (y + 1)² = -iv(x - 9/4). From hither nosotros can say that the parabola will open to the left.

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

The term in front of the 10 is a -four. This is our 4p value. And then nosotros now can say that 4p = -4. In turn, our p = -1.

Now will we determine that the focus is i unit of measurement left of the vertex; so the focus (after some work with fractions) is (v/4, -ane).

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

Beneath, nosotros will encounter the sketch of this equation.

Equally has been the case so far, the plus sign represents the focus, which resides at the point (5/4, -1); the directrix is represented by the green line, which is on the equation x = xiii/iv.

For some supplementary exercises over what we accept covered in lesson four, click here: Exercise 4

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